consider the following hypothesis test:h0: m525 ha: m ±25 a sample of 40 provided a sample mean of 26.4. the population standard deviation is 6.a. compute the value of the test statistic.b. what is the p-value ?c. at a = .01, what is your conclusion?d. what is the rejection rule using the critical value? what is your conclusion?
Question
consider the following hypothesis test:h0: m525 ha: m ±25 a sample of 40 provided a sample mean of 26.4. the population standard deviation is 6.a. compute the value of the test statistic.b. what is the p-value ?c. at a = .01, what is your conclusion?d. what is the rejection rule using the critical value? what is your conclusion?
Solution
a. The test statistic is calculated using the formula:
Z = (X - μ) / (σ/√n)
where: X = sample mean = 26.4 μ = population mean = 25 σ = population standard deviation = 6 n = sample size = 40
Substituting these values into the formula, we get:
Z = (26.4 - 25) / (6/√40) = 1.4
b. The p-value is the probability that you would observe a result as extreme as the test statistic (or more extreme) if the null hypothesis is true. To find the p-value, we look up the test statistic (1.4) in a standard normal (Z) table or use a Z-score calculator. The p-value is approximately 0.08.
c. The significance level (α) is 0.01. Since the p-value (0.08) is greater than the significance level, we do not reject the null hypothesis. This means that we do not have enough evidence to support the alternative hypothesis.
d. The rejection rule using the critical value is that if the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. The critical value for a two-tailed test at a significance level of 0.01 is approximately 2.58. Since the absolute value of our test statistic (1.4) is less than the critical value, we do not reject the null hypothesis. This conclusion is consistent with our conclusion from part c.
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