Question

This question requires coding in RStudio. It is from the book Verzani- Problem 9.9:

The United States Department of Energy conducts weekly phone surveys on the price of gasoline sold in the United States. Suppose one week the sample average was $4.03, the sample standard deviation was $0.42, and the sample size was 800. Perform a one-sided significance test of H₀ : $\mu$ = 4.00 against the alternative H_A : $\mu$ > 4.00.

Answer 1

Given : Sample size=n=800

Sample mean= $\bar{X}=4.03$

Sample standard deviation=s=0.42

Hypothesis : $H_0:\mu=4.00$ Vs $H_A:\mu>4.00$

The test statistic is ,

$Z=\frac{\bar{X}-\mu_0}{s/\sqrt{n}}\to N(0,1)$

$=\frac{4.03-4.00}{0.42/\sqrt{800}}=2.02$

Critical value : $Z_{\alpha}=Z_{0.05}=1.64$

Decision : Here , $Z_{stat}=2.02>Z_{\alpha}=1.64$

Therefore , reject the null hypothesis at 0.05 Significance level

Conclusion : Hence , there is sufficient evidence to conclude that the population mean is greater than 4.00.