7.2 Hypothesis testing for the mean (sigma known)

Find the​ P-value for a​ left-tailed hypothesis test with a test statistic of Z=−1.15. Decide whether to reject H0 if the level of significance is α=0.05

area to the left of z= normalcdf(-10,000, -1.15,0,1)= 0.1251 p-value= 0.1251 To use a​ P-value to make a conclusion in a hypothesis​ test, compare the​ P-value with alphaα. If P≤ α​, then reject H0. If P>α​, then fail to reject H0. Since P>α​, fail to reject H0.

Explain the difference between the​ z-test for μ using rejection​ region(s) and the​ z-test for μ using a​ P-value.

In the​ z-test using rejection​ region(s), the test statistic is compared with critical values. The​ z-test using a​ P-value compares the​ P-value with the level of significance α. A rejection region​ (or critical​ region) of the sampling distribution is the range of values for which the null hypothesis is not probable. A critical value z0 separates the rejection region from the nonrejection region. To use a rejection region to conduct a​ z-test, calculate the standardized test statistic z. If the standardized test statistic is in the rejection​ region, then reject H0. If the standardized test statistic is not in the rejection​ region, then fail to reject H0. To use a​ P-value to make a conclusion in a hypothesis​ test, compare the​ P-value with α. If P ≤ α​, then reject H0. If P > α​, then fail to reject H0.

Find the​ P-value for the indicated hypothesis test with the given standardized test​ statistic, z. Decide whether to reject H0 for the given level of significance α. ​Two-tailed test with test statistic z=−2.18 and α=0.02

Since this is a​ two-tailed test and the test statistic is left of​ center, the​ P-value is twice the area to the left of the test statistic z-score-= normalcdf(-10000,-2.18,0,1) multiplied by 2 to find p-value= 0.0292

Find the critical​ value(s) and rejection​ region(s) for the type of​ z-test with level of significance α. Include a graph with your answer. Right​-tailed ​test, α=0.10

The critical values are z=1.28 (1-alpha= 0.9; invNorm(0.9,0,1)= 1.28 The rejection region is z >1.28 Pick the right-tailed graph

State whether the standardized test statistic z indicates that you should reject the null hypothesis. (left-tailed) ​(a) z=1.208 ​(b) z=−1.364 ​(c) z=−1.467 ​(d) z=- 1.189 a) For z=1.208, should you reject or fail to reject the null​ hypothesis?

a) Fail to reject H0 because z > −1.285. b)Reject H0 because z< −1.285. c) Reject H0 because z<−1.285. d)Fail to reject H0 becuase z > -1.285

A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 750 hours. A random sample of 24 light bulbs has a mean life of 728 hours. Assume the population is normally distributed and the population standard deviation is 65 hours. At α=0.02​, do you have enough evidence to reject the​ manufacturer's claim? This is left-tailed. a)) Identify the null hypothesis and alternative hypothesis. b) Identify the critical​ value(s). C) Identify the standardized test statistic d) Decide whether to reject or fail to reject the null​ hypothesis (e) interpret the decision in the context of the original claim.

a) H0: mu equal than > 750 (claim) Ha: mu < 750 b) null z: invNorm(alpha,0,1)= -2.05 c)z= -1.66 (used STAT tests menu, p. 368) d) Fail to reject H0. There is not sufficient evidence to reject the claim that mean bulb life is at least 750 hours. e)

Determine whether to reject or fail to reject H0 at the level of significance of​ a)α=0.07 and​ b) α=0.02. H0​: μ=123​, Ha​: μ≠123, and P=0.0396

a) Reject H0 because P<0.07 b) Fail to reject H0 because P>0.02

Find the​ P-value for the indicated hypothesis test with the given standardized test​ statistic, z. Decide whether to reject H0 for the given level of significance α. ​ Right-tailed test with test statistic z=1.29 and α=0.04

normalcdf(1.29,10,000,0,1) p-value= 0.0985 Fail to reject H0 cuz p-value is higher than alpha

In hypothesis​ testing, does choosing between the critical value method or the​ P-value method affect your​ conclusion?

​No, because both involve comparing the test​ statistic's probability with the level of significance The​ P-value method converts the standardized test statistic to a probability​ (P-value) and compares this with the level of​ significance, whereas the critical value method converts the level of significance to a​ z-score and compares this with the standardized test statistic.​ Thus, both methods will result in the same conclusion.