belindajolete
belindajolete
26.02.2020 • 
Mathematics

A confidence interval for the population mean tells us which values of are plausible (those inside the interval) and which values are not plausible (those outside the interval) at the chosen level of confidence. You can use this idea to carry out a test of any null hypothesis H0:=0 and reject H0 if 0 is outside the interval and fail to reject H0 if 0 is inside the interval. The alternative hypothesis is always two‑sided, H:≠0, because the confidence interval extends in both directions from x¯. A 95% confidence interval leads to a test at the 5% significance level because the interval is wrong 5% of the time. In general, confidence level leads to a test at significance level α=1−. (

a. A medical director found mean blood pressure x¯=128.07 for an SRS of 72 male executives between the ages of 50 and 59. The standard deviation of the blood pressures of all males 50–59 years of age is σ=15. Select the 90% confidence interval for the mean blood pressure of all executives in this age group, assuming the standard deviation is the same for all males 50‐59 years of age. 125.80 to 130.34 125.16 to 130.98 124.61 to 131.53 123.52 to 132.62
b. The hypothesized value 0=130 falls inside this confidence interval. Carry out the z test for H0:=130 against the two‑sided alternative. Find the test statistic z. (Enter your answer rounded to two decimal places.) z=

Find the P ‑value for the z test
P ‑value =

Select all the statements that explain why the test is not statistically significant at the 10% level.
i. Because the hypothesized mean falls inside the 90% confidence interval.
ii. Because the P ‑value is greater than 0.1. Because the true mean falls inside the 90% confidence interval.
iii. Because the margin of error for a 90% confidence interval is less than 10% of the hypothesized mean.

(c) The hypothesized value 0=131 falls outside this confidence interval. Carry out the z test for H0:=131 against the two‑sided alternative.

Find the test statistic z.
Find the P ‑value for the z test using
P ‑value =

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