Question

If X-bar= 95, S = 22, and n = 64, and assuming that the population is normally distributed,

a. Construct a 99% confidence interval for the population mean, μ.

b. Based on your answer to part (a), test the null hypothesis that the population mean μ = 101 vs. the alternative that μ ≠ 101.

c. What is the probability that μ = 101? d. What is the probability that μ > 101?

Answer 1

Solution:-

a) 99% confidence interval for the population mean, μ is C.I = ( 87.916, 102.084).

C.I = 95 + 2.576*2.75

C.I = 95 + 7.084

C.I = ( 87.916, 102.084)

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 101
Alternative hypothesis: u 101

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 2.75
z = (x - u) / SE

z = - 2.18

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z statistic less than -2.18 or greater than 2.18.

Thus, the P-value = 0.029.

Interpret results. Since the P-value (0.029) is greater than the significance level (0.01), we cannot reject the null hypothesis.

c) The probability that μ = 101 is 0.0134

x = 101

By applying normal distribution:-

z = 2.1818

P(z = 2.1818) = 0.0134

d) The probability that μ > 101 is 0.0146

x = 101

By applying normal distribution:-

z = 2.1818

P(z > 1.8) = 0.0146