Question

5c. Suppose you have the following information: i = 64500, n = 30, and a = 800 Complete the graph of the sampling distribution of the test statistic and conclude. Do you Reject or Fail to Reject H? - Step 1: H: = 65,000 H: #65,000 - Step 2: a = 0.05 - Step 3: Step 4 (critical value): - Step 5: Standard Normal Distribution (2) 0.4 0.2 TTTTT 0.0 w

Answer 1

Sample size = n = 30

Sample mean = $\dpi{80} \overline{X}$ = 64,500

Population standard deviation = $\dpi{100} \sigma _{x}$ = 800

H₀: $\dpi{100} \mu$ = 65,000

H_a: $\dpi{100} \mu$$\dpi{100} \neq$ 65,000

Level of significance = $\dpi{100} \alpha$ = 5% = 0.05

Since sample size is 30 ( greater than or equal to 30) and population standard deviation is given, we can use standard normal distribution for the sampling distribution of mean.

Since, it's a two-sided hypothesis test, we would check for the values of (level of significance / 2) = 0.05/2 = 0.025.

At this level of significance, using Z - table, critical values for 0.025 are -1.96 and 1.96

test statistic = ($\dpi{80} \overline{X}$ - $\dpi{100} \mu$) / SE(X)

where SE(X) is the standard error of $\dpi{80} \overline{X}$ , which is given by, SE(X) = $\dpi{100} \sigma _{x}$ /$\dpi{80} \sqrt{n}$

So, test statistic = ($\dpi{80} \overline{X}$ - $\dpi{100} \mu$) /  ($\dpi{100} \sigma _{x}$/$\dpi{80} \sqrt{n}$)

= (64,500 - 65,000) / (800/$\dpi{80} \sqrt{30}$)

= -3.4232

Since -3.4232 is outside the range of -1.96 and 1.96, we would reject the null hypothesis.