Unit 5

Standard Error for Sample Proportions (Population SD unknown)

(phat)(qhat) dived by n sqaured p hat = sample proportion of success (said yes) q hat = complement of p hat (1- p hat) n = sample size

Jason sampled 60 smokers who were questioned about the number of hours they sleep each day. Jason wants to test the hypothesis that the smokers need less sleep than the general public which needs an average of 7.5 hours of sleep with a population standard deviation of 0.7 hours.If the sample mean is 7.2 hours, what is the z-score? Answers are rounded to the hundredths place.

-3.32 z = x-u divdied by o/square root of n. x = Sample Mean U = Population Mean O= SD N = sample Size

Julie knows that the adult population gets, on average, eight hours of sleep each night. A hypothesis test can help her see if college students are different from the adult population.Julie tabulated that her sample of 101 students got an average of 7.1 hours of sleep each night, with a standard deviation of 2.48.Using the data provided and the formula below, what is the t-statistic that Julie calculates? Answer choices are rounded to the hundredths place.

-3.46 x = 7.1 u = 8 s - 2.48 n = 101

The average weight of a cereal box is 25 ounces. William believes the cereal boxes at his local grocery store are less than the claimed average and decided to sample 15 cereal boxes from this grocery store.Using the alternative hypothesis that µ < 25, William found a t-test statistic of -2.624.What is the p-value of the test statistic? Answer choices are rounded to the hundredths place.

0.01 In order to find the p-value, we first need the degrees of freedom because we are running a t-test. The df = n-1 = 15-1 = 14. Since this is a one-tailed test, in order to find the p-value we simply need to find the probability of being as extreme or more extreme as our current test statistic. Another way to describe this is it is the tail probability/area. If we go to the df row of 14 and then scan to the right we see 2.624 is in the 0.01 column. This indicates there is 0.01 or 1% in the tail. So the p-value is 0.01.

An article claims that 12% of trees are infested by a bark beetle. A random sample of 1,000 trees were tested for traces of the infestation and found that 127 trees were affected.Using the formula and data provided, what is the value of the z-test statistic? Answer choices are rounded to the hundredths place.

0.70 p hat = 127/1000 = 0.127 p = 0.12 q= 1-0.12 = 0.88 n= 1000

Cindy recorded the number of customers who visited her new shop during the week: DayCustomersWednesday18Thursday20Friday25Saturday17 She expected to have 20 customers each day. To determine whether the number of customers follows a uniform distribution, a chi-square test for goodness of fit should be performed (alpha = 0.10).

1.9

Marcus collected 30 responses to a survey and used a two-tailed test to establish a significance level of 0.05 and calculate a 95% confidence interval.Using the t-table (located in the tutorial), what is the critical t-value used for calculating a 95% confidence interval with 29 degrees of freedom? Answer choices are rounded to the thousandths place

2.045 If the sample is n = 30, then the df = n-1 = 29. Recall that a 95% confidence interval would have 5% for the tails, so 2.5% for each tail. So if we go to the row where df = 29 and then 0.025 for the tail probability, this gives us a value of 2.045. We can also use the last row and find the corresponding confidence level.

What value of z* should be used to construct a 99% confidence interval of a population mean? Answer choices are rounded to the hundredths place.

2.58 Using the z-chart to construct a 99% CI, this means that there is 1% for both tails, or 1% / 2 = 0.5% or 0.005 for each tail. The lower tail would be at 0.005 and the upper tail would be at (1 - 0.005) or 0.995.The closest to 0.995 IN THE Z-TABLE is at 0.9951, which corresponds with a z-score of 2.58.

If the average winter temperature in Miami is 69°F with a standard deviation of 7°F, what percentage of days would the temperature be above 73°F? Answers are rounded to the nearest whole percent.

28% 73-69 divided by 7 = 4/7 = 0.57. On the z table a score of 0.57 corresponds with a percentage of 0.74 or about 0.72 = 72%. However this is the percentage for the lower distribution. We want the value for the upper, so we subtract this from 100%. 100-72 = 28.

A table represents the possibility of an association between hair color and eye color. Eye ColorHair ColorBlueGreenBrownBlonde252731Brown261822 In order to determine if the eye color of blue and blonde hair colors differ significantly, a chi-square test for homogeneity should be performed.What is the expected frequency of Blonde Hair and Blue Eyes? Answer choices are rounded to the nearest hundredth.

28.41 E = (row total)(colum total) divided by the complete total. Blonde hair is 83 Blue eyes is 51 Whole table is 149 83*51 divided by 149 = 28.41

Jackson sampled 101 students and calculated an average of 6.5 hours of sleep each night with a standard deviation of 2.14. Using a 95% confidence level, he also found that t* = 1.984.A 95% confidence interval calculates that the average number of hours of sleep for working college students is between __________ hours. Answer choices are rounded to the hundredths place.

6.08 and 6.92 x = 6.5 t* = 1.984 s = 2.14 n = 101

A school authority claims that the average height of students is 178 cm. A researcher has taken a well-designed survey and his sample mean is 177.5 cm and the sample standard deviation is 2. The sample size is 25.Which statement is correct?

A a.) The difference exists due to chance since the test statistic is small. b.) The sample mean and population mean is the same. c.) The result of the survey is statistically significant. d.) The result of the survey is biased. With a very small sample size of 25, a difference of 0.5 cm is most likely due to chance.

A research company is claiming that 75% of college students shop online. Jamila takes a sample of 200 college students and finds that 120 college students shop online.Which type of inference test should Jamila use?

A a.)One proportion z-test b.)Chi-squared test for homogeneity c.)Two-way ANOVA d.)One-sample t-test Since we are testing a proportion from one group using a sample that is of sufficient size, we would use a one proportion z-test.

Null Hypothesis

A claim about a particular value of a population parameter that serves as the starting assumption for a hypothesis test. Ho

Alternative Hypothesis

A claim that a population parameter differs from the value claimed in the null hypothesis. Ha

Sample Mean

A mean obtained from a sample of a given size. Denoted as X(line over it)

One tailed Test

A test for when you have a reason to believe the population parameter is higher or lower than the assumed parameter value of the null hypothesis. Right Tailed Left Tailed Only test whether or not there is evidence of a statistic being significantly higher or lower than a claimed parameter., like mu or p.

Which of the following statements is FALSE? Between significance level and the power of a hypothesis test.

A. a.) Expanding the sample size can decrease the power of a hypothesis test. b.) The probability of rejecting the null hypothesis in error is called a Type I Error. c.) A larger sample size would increase the power of a significance test. d.) The significance level is the probability of making a Type I error. If you increase the sample size, the ability to detect differences increases, which reduces Type II error. So this means the power of the test has increased.

Which of the following statements is FALSE? Between Significance level and the power of a hypothesis test?

A. a.) Reducing the significance level (α) can increase a test's effectiveness. b.) Alpha (α) is equal to the probability of making a Type I error. c.) ​Expanding the sample size can increase the power of a hypothesis test. d.) A larger sample size would increase the effectiveness of a hypothesis test Recall that alpha is Type I error and corresponds to the threshold we use of acceptable Type I error. So if we increase alpha we are opening ourselves up to more Type I error. This is not an increase in the test's effectiveness.

Analysis of Variance (ANOVA)

A: Analysis O: of VA: Variance Compare the means by analyzing the sample variances from the independently selected sample. Hypothesis test that allows us to compare three or more population means. Conditions: Independent samples from the populations. Each population has to be normally distributed The variance and therefore the standard deviations of all those normal distributions are the same.

Standard Deviation of Sampling Distribution of Sample Means

Also called the standard error. IT's the standard deviation of the population, divided by the square root of the sample size.

Type II Error

An error that occurs when a false null hypothesis is not rejected.

A school authority claims that the average percentage marks of students is 68. A researcher has taken a well-designed survey and his sample mean is 64.5 and sample standard deviation is 2. The sample size is 300.Which statement is correct?

B a.) The difference exists due to chance since the test statistic is small. b.) The result of the survey is statistically significant. c.) The sample size should be much more. d.) The result of the survey is not statistically significant. Since we find that 64.5 is much lower than the average 68 from a large sample size and relatively small standard deviation, we can conclude there is a statistically significant result. It is also practically different as well.

A recent article claims that the state of Illinois has low tuition rates for its colleges and universities. Cate decides to research the cost of tuition for different colleges in Illinois. The average tuition in Illinois is $28,950 with a standard deviation of $3,470.Which type of inference test should Cate use?

B a.)Two-way ANOVA b.)One-sample z-test c.)Chi-squared test for goodness-of-fit d.)One-sample t-test Assuming we have a large enough sample, since we are testing one group's mean relative to a hypothesized value, this would be a one-sample z-test.

Miriam is using a one-sample t-test on the following group:Subject #15: 6.5 hoursSubject #27: 5 hoursSubject #48: 6 hoursSubject #80: 7.5 hoursSubject #91: 5.5 hoursSelect the two TRUE statements.

B & C. a.)Miriam needs to use a t-test when the standard deviation is known. b.)The value for the degrees of freedom for Miriam's sample population is four. c.)The t-distribution that Miriam uses is shorter and has thicker tails than a normal distribution. d.)The value for the degrees of freedom for Miriam's sample population is five. e.)The t-distribution that Miriam uses is taller and has thinner tails than a normal distribution. For the t-distribution, we use the degrees of freedom to find the critical value and the degrees of freedom are n-1. So in this case n =5 and df = 5-1 = 4. Also, recall that the t-distribution for lower degrees of freedom has shorter, fatter tails than the standard normal distribution.

If the average cholesterol level is 194 with a standard deviation of 15, what percentage of children have a cholesterol level lower than 199? Answers are rounded to the nearest whole percent.

B. a.)37% b.)63% c.)74% d.)26% Recall to find the probability, we need to convert to a z-score and then go to standard normal chart. We can note: z= value-mean divided by SD = 0.33. Looking at a Z-table, a z-score of 0.33 corresponds with a percentage of 0.6293 or about 0.63.

Mark is using a one-sample t-test on the following group:Subject #15: 7.5 hoursSubject #27: 6 hoursSubject #48: 7 hoursSubject #80: 6.5 hoursSubject #91: 6.5 hoursSubject #59: 7 hoursSelect the two TRUE statements.

C & D a.)The t-distribution that Mark uses has thinner tails than a standard distribution. b.)Mark would use the population standard deviation to calculate a t-distribution. c.)The value for the degrees of freedom for Mark's sample population is five. d.)The t-distribution that Mark uses has thicker tails than a standard normal distribution. e.)The value for the degrees of freedom for Mark's sample population is six. For the t-distribution, we use the degrees of freedom to find the critical value and the degrees of freedom are n-1. So in this case n = 6 and df = 6-1 = 5. Also, recall that the t-distribution for lower degrees of freedom has shorter, fatter tails than the standard normal distribution.

Theresa really loves orange-flavored Fruity Tooty candies, but there always seems to be a lot of cherry-flavored candies in each bag. To determine whether this is because cherry candies are so popular or because each bag contains fewer orange candies, Theresa randomly picks a Fruity Tooty candy from her bag, records its flavor, and places it back in the bag. Each bag contains a mixture of cherry, grape, apple, lemon, and orange flavors.Which statement about Theresa's distribution of sample proportions is true?

C. a.)The count of drawing an orange candy is not a binomial distribution. (Not this because this would have two categories, orange and not orange.) b.) The distribution of lemon candy can be modeled as normal if Theresa picks candies 12 times. c.) The sample proportion of cherry candies has a binomial distribution. d.) The distribution of the count of picking an apple candy cannot be modeled as approximately normal if Theresa picks candies over 300 times. Recall proportions can be put into two categories. IF we assume that items are independently chosen, this would follow the binomial distribution. She can either pick a cherry candy or not pick a cherry candy.

F-Statistic

Calculated by taking the quotient of the variability between the samples and the variability within each sample. F = variability between the samples divided by variability within each sample

Confidence Interval for Population Proportions

Confidence interval is an estimate found by using a sample statistic and adding and subtracting an amount corresponding to how confident we are that the interval created captures the population parameter. Statistic is the same proportion and population parameter is the population proportion.

Two way ANOVA

Consider the population means based on multiple characteristics. -Comparing 3 more sample means for multiple characteristics. Example:

One way ANOVA

Consider the population means based on one characteristic - Comparing 3 or more sample means for only one characteristic. Example: Studying the number of hours of sleep for teenagers, adults, and senior citizens.

Population Parameter

Corresponding measurement for the population. Something that we can find in a sample. The only way to figure out a parameter is to take a census.

Calculate a critical z-score for a left tailed, right tailed or two tailed test. Which statement is correct?

D a.) The critical z-score for a two-sided test at a 4% significance level is 1.75. b.)The critical z-score for a two-sided test at a 20% significance level is 0.85. c.)The critical z-score for a left-tailed test at a 12% significance level is -0.45. d.)The critical z-score for a right-tailed test at a 9% significance level is 1.34. For a right-tailed test, we want to find the value in the upper tail. To do this, we will subtract the significance level from 100%: 100% - 9% = 91% or 0.91.In a z-table, the probability of 0.91 is closest to 0.9099 which corresponds to a z-score of 1.34.

Select the correct statement. Determine whether to reject a null hypothesis from a given p value and significance level.

D a.)Given a p-value of 0.05, and a significance level of 3%, you should reject the null hypothesis.​ b.)Given a p-value of 0.08, and a significance level of 2%, you should reject the null hypothesis.​ c.)Given a p-value of 0.06, and a significance level of 5%, you should reject the null hypothesis.​ d.)Given a p-value of 0.01, and a significance level of 5%, you should reject the null hypothesis.​ Recall that our decision rule is to reject Ho if p-value is less than significance level. Since we have 0.01 (p-value) is less than 0.05 (significance level), then we should reject Ho.

A spice box manufacturing company is having difficulty filling packets to the required 50 grams. Suppose a business researcher randomly selects 60 boxes, weighs each of them and computes its mean. By chance, the researcher selects packets that have been filled adequately and that is how he gets the mean weight of 50 g, which falls in the "fail to reject" region.The decision is to fail to reject the null hypothesis even though population mean is NOT actually 50 g.Which kind of error has the researcher done in this case?

D a.)Neither b.)Type I c.)Both d.)Type II Since the sample evidence provides evidence the null should not be rejected, when in fact the null hypothesis is false, this is called Type II error.

Select the true statement between sample size and the standard deviation of distribution of sample means.

D a.)Sample size does not have an impact on standard error. b.)As sample size increases, standard error increases. c.)As sample size decreases, standard error decreases. d.)As sample size increases, standard error decreases.

Jeanette really loves apple-flavored Fruity Tooty candies, but there always seems to be a lot of cherry-flavored candies in each bag. To determine whether this is because cherry candies are so popular or because each bag contains fewer apple candies, Jeanette randomly picks a Fruity Tooty candy from her bag, records its flavor, and places it back in the bag. Each bag contains a mixture of cherry, grape, apple, lemon, and orange flavors. Which statement about Jeanette's distribution of sample proportions is true?

D a.)The distribution of the count of picking a cherry candy cannot be modeled as approximately normal if Jeanette picks candies over 100 times. b.) The distribution of the count of picking a lemon candy can be modeled as approximately normal if Jeanette picks candies 15 times. c.) The count of drawing an apple candy is not a binomial distribution. d.)The count of drawing an orange candy has a binomial distribution. Recall that proportions can be put into two categories. If we assume that items are independently chosen, this would follow the binomial distribution.

Susan really loves the lemon-flavored Fruity Tooty candies, but there always seems to be a lot of grape-flavored candies in each bag. To determine whether this is because grape candies are so popular or because each bag contains fewer lemon candies, Susan randomly picks a Fruity Tooty candy from her bag, records its flavor, and places it back in the bag. Each bag contains a mixture of cherry, grape, apple, lemon, and orange flavors.Which statement about Susan's distribution of sample proportions is true?

D. a.) The distribution of the count of picking a cherry candy cannot be modeled as approximately normal if Susan picks candies over 200 times. b.)The sample proportion of drawing an orange candy is not a binomial distribution. c.) The distribution of lemon candy can be modeled as normal if Susan picks candies 10 times. d.) The distribution of apple candy can be modeled as normal if Susan picks candies over 75 times. In General the sampling distribution of the proportions follows a binomial distribution. As you let the sample size get larger, we can use a normal approximation to the binomial and note that the sampling distribution converges to normal distribution.

Degrees of Freedom

Degree's of freedom = (row total - 1)(Column total - 1)

Population Mean

Denoted as the greek letter mu Population proportion = p Population Mean=. greek U Population Std Dev = fancy o

Distribution of Sample Proportions

Distribution of all possible sample proportions for a certain size, n. A distribution where each data point consists of a proportion of successes of a collected sample. For a given sample size, every possible sample proportion will be plotted in the distribution.

Distribution of Sample Means

Distribution that shows the means from all possible samplings of a given size. Each data points consist of a mean of a collected sample. Every possible sample mean will be plotted in the distribution.

Type 1 Error

Error that occurs when a true null hypothesis is rejected.

Find the P Value with the Z Score

Excel: Formula -> Statistical -> NORM.DIST. Input th emean of the sample., then the population mean, then the standard deviation, divided by the squre root of the sample n, enter true) Example: =NORM.DIST(SAMPMEAN, POPMEAN,SD/SQRT(SAMPLE POPULATION), TRUE)

Tailed Test on Excel

Formulas - Statisticals - Norm.S.INV - enter in decimal point, hit enter.

Chi-Squared Test for Goodness of Fit

H0 = Population Distribution matches a specified distribution Ha = Population distribution does not match a specified distribution.

Hint

If p is the probability of success, q is the probability of failure, which is equal to 1-p.

Confidence Interval

Interval that contains the likely values for a parameter. We base the confidence interval on our point estimate, and the width of the interval is affected by the confidence level and sample size. CI= Point Estimate + or - Margin of Error

From her purchased bags, Rachel counted 130 red candies out of 520 total candies.Using a 95% confidence interval for the population proportion, what are the lower and upper limit of the interval? Answer choices are rounded to the thousandths place.

Lower Limit: 0.213 Upper Limit: 0.287 To calculate the limits of a confidence interval, we first need to find the z-score for 95% CI. Recall that 95% CI means there is 5% for both tails, or 2.5% in each tail. This relates to a z-score of 1.96.Confidence interval can be found using the formula . Plug in the following values: p hat = 130/520 =0.25 q hat = 1- p hat = 1 - 0.25 = 0.75 z star = z-score for 90% CI = 1.96 N = 520 0.25 plus or minus 1.96 times the square root of (0.25)(0.75) divided by 520

Sample Statistic

Measure of an attribute of a sample. Sample proportion = P(hat) Sample Mean = X (Line over it) Sample Std Dev = s sample mean is the sum of a certain attribute of a sample divided by the sample size.

T-Distribution

More heavy tailed. A family of distributions, a little shorter than the standard normal distribution and a little heavier on the tails. Sample size gets larger, the t-distribution does get close to the normal distribution. family of distributions that are centered at zero and symmetric like the standard normal distribution, but heavier in the tails. Does not diminish sh towards the tails as fast.

Observed Frequency

Number of observations we actually see for value, or what actually happened.

Proportion of Successes

Number of successes divided by the sample size, n. The Standard deviation will be the standard deviation of the binomial distribution, divided by n.

The Fruity Tooty company claims that the population proportion for each of its five flavors is exactly 20%. Jane counts 92 red candies in a 400-count sample.Using the formula and data provided, what is the value of the z-test statistic? Answer choices are rounded to the hundredths place.

P hat = 92/400 = 0.23 p = 0.20 q = 1-p = 1-0.20 = 0.80 n = 400

Chi-Square Statistic

Particular Test used for categorical data. Measures how expected frequency differs from observed frequency. Formula: Observed values, - expected values, square the different, divid by the expected values and then add them up. x squared = Observed - Expected Squared divided by Expected Smaller: Indicate a small discrepancy Larger: Large Discrepancy

Power of a Hypothesis Test

Probability of rejecting the null hypothesis correctly, rejecting when the null hypothesis is false, which is a correct decision. Probability that we reject the null hypothesis when a difference truly does exist.

Test Statistic

Relative distance of the statistic obtained from the sample from hypothesized value of the parameter from the null hypothesis. Measured in terms of the number of standard deviations from the mean of the sampling distribution. Z-Statistic or Z-Score Measurement in standardized units of how far a sample statistic is from the assumed parameter if the null hypothesis is true.

Margin of Error and the Affect of Confidence Level and Sample Size

Sample Size: Larger sample size results in less sampling error, and therefore a lower margin of error. Confidence Level: Higher confidence level results in a larger margin of error.

Standard Deviation of a Distribution of Sample Proportions

Square root of n times p times q, divded by n. Also known as the standard error. A measure calculated by taking the square root of the quotient of p(1-p) and n.

T - Test for Population Means

Standard deviation Sample, s. t = sample mean minus the hypothesized population mean, over the standard error dived by the square root of the sample size. Problem: The value of s can vary largely from sample to same. For large samples, s and sigma are very close, but with small samples particularly, the value of s can vary. Because it varies so much , they created a new distribution of test statistics much like the z test, but now it's t distribution.

Standard Error

Standard deviation of the sampling distribution of sample means

Hypothesis tsting

Standard procedure in statistics for testing a hypothesis, or claim about population parameters.

Test Statistic Formula

Statistic - parameter \ standard deviation of statistic

Z-Statistic of Means

Statistic : Sample Mean X_, Parameter: Hypothesized population mean: u, Standard Deviation of x_: o divided by square root of n. Formula: x-u divided by o that is divided by the square root of n.

Z Test

Statistic is equal to the sample mean minus the hypothesized population mean, over the standard deviation of the population dived by the square root of sample size. Used if the Population SD is know

Z-Statistic for Proportions

Statistic: Sample proportion: phat Parameter: Hypothesized population proportion: p Standard Deviation: Phat: quare root of pq divided by n. The standard deviation of the phat statistic is going to be the square root of p times q which is 1-p, over n. Therefore, the z-statistic for sample proportions that you can calculate is your test statistic, and it is equal to phat minus p from the null hypothesis, divided by the standard deviation of phat.

Represent all the distrubtions

Step 1: Take the sample means and graph them. Draw out an axis. Step 2: Take the average value, for example, the mean of 2.5, and put a dot at 2.5 on the x-axis much like a dot plot. Do this for all the sample means that you have found. Step 3: Keep doing this over and over again. Ideally you would do this hundreds or thousands of times to show the distribution of all possible samples that could be taken.

Standard Normal Table

Table that is used when you have a normal distribution and you want to find probabilities or percent. Used for: 1.The table value itself gives you the percent of observations below a particular z-score. 2.You can find the percent above a particular z-score by subtracting the table value from 100% because the table value always gives the area to the left. 3.You can find the percent of observations between two z-scores by subtracting the table values. 4.You can find the percent of values outside of two z-scores by finding both the percent above the higher number and the percent below the lower number, which is sort of a combination of these other options.

Proportions

The only way to obtain the true population proportion, which is the parameter we're trying to estimate, is by taking a cenus.

Significance Level

The probability of making a Type 1 Error. Abbreviated with the symbol alpha, a. Being comfortable making some error sometimes.

Select the False statement about sampling error and sample size

The standard error increases as the sample size increase. True: The sample size can impact the sampling error The larger a sample size, the more accurate an estimate can be In order to decrease sampling error, you must increase the sampling size

Left Tailed Test

Type of One tailed test in which alternative hypothesis claims that it's less than the claimed parameter. Looking at the values lower than the assumed value, which is the section to the left of the value.

Right Tailed Test

Type of one tailed test that means that the alternative hypothesis is larger than the claimed parameter. The distribution of a right tailed test would look similar to the following curve.

Chi-Square Test of Homogeneity

Uses Multiple populations and tests to see if these populations are the same across categorical or qualitative, variables. Trying to determine if the distributions of categorical data differs across different populations.

Mean of Distribution of Sample Proportions

Value of P, which is the actual probability.

Expected Frequency

What we would expect to happen. The number of observations we would see for a value if the null hypothesis was true.

Two Tailed Test

When we have no reason to believe the population is different from the assumed parameter value of the null hypothesis. Looking at the values on the values that are extremely lower or higher than the assumed value. Not equal to.

Parameters

a numerical value that characterizes some aspect of a population. Example: Average number of all adults who smoke regularly Symbols: u , o

Sampling with replacement

means that you put everything back once you've selected it. Typically, one big requirement for statistical inference is that the individuals, the values from the sample, are independent. One doesn't affect any of the others. When sampling with replacement, each trial is independent.

DEGREE OF FREEDOM

n-1

Standard Error of a Sample Mean

s / square root of n Used when we have quantitative data and can be calculated used the following formula.

Statistics

sample measures that we can use to estimate parameters, which are corresponding population measure. It's important to remember that this only works when the sample is carried out well. For instance, if there's bias, then those wouldn't accurately reflect the population measures.

Sampling Error

simply relates to the variability within the sampling distribution. It is the amount by which the sample statistic differs from the population parameter. Standard deviation decrease as the sample size increases.

sampling without replacement

which means that each observation is not put back once it's selected--once it's selected, it's out and cannot be selected again. For independence, a large population is going to be at least 10 times larger than the sample.