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A) What is the standard error of the mean for the sample data provided: sample size = 64, sample mean = 17.5, sample standard deviation = 2.2?

b) What is the critical value of t* for a 95% confidence interval?

c) Calculate and construct a 95% confidence interval for the population mean (μ) based on the sample data.

d) At a significance level of α = 0.025, can you reject the null hypothesis H0: μ = 18 in favor of the alternative hypothesis HA: μ < 18, based on the confidence interval? Please explain your decision.

User Gushitong
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Final answer:

The standard error of the mean is 0.275. The 95% confidence interval for the population mean is (16.95, 18.05). Since 18 is within this range, we cannot reject the null hypothesis H0: μ = 18 at the α = 0.025 significance level.

Step-by-step explanation:

To calculate the standard error of the mean for the provided sample data, we use the formula SE = s / √n, where 's' is the sample standard deviation, and 'n' is the sample size. Plugging in, we get SE = 2.2 / √64 = 2.2 / 8 = 0.275.

For a 95% confidence interval, the critical value of t* (t-star) depends on the degrees of freedom (df), which is n - 1 for a sample for a t-distribution. So df = 64 - 1 = 63. Using a t-distribution table or calculator, we find the t* value for df = 63 and a two-sided 95% confidence interval, which is approximately 2.00.

To construct the 95% confidence interval for the population mean (μ), we use the formula: mean ± t* * SE. Substituting the values, we get: 17.5 ± 2.00 * 0.275, which equals to 17.5 ± 0.55, hence the interval is (16.95, 18.05).

Considering a significance level of α = 0.025, whether we can reject the null hypothesis H0: μ = 18 depends on if 18 falls outside the constructed confidence interval. Since 18 is inside the interval (16.95, 18.05), we cannot reject the null hypothesis; thus, we don't have sufficient evidence to support the alternative hypothesis HA: μ < 18.

User Yanokwa
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