Can someone please help how to solve this I keep getting the wrong answer so I am now freaking out and clueless
23) The following data show the average retirement ages for a random sample of workers in Boscov’s and a random sample of workers in ShopRite.
Boscov’s: Sample mean is 64.6 years, sample size is 30, population standard deviation is 4 years
ShopRite: Sample mean is 67.5 years, sample size is 30, population standard deviation is 4.5 years
a) Perform a hypothesis test using alpha = .05 to determine if the average retirement age at ShopRite is higher than it is in Boscov’s.
b) Determine the p value and interpret the results.
a)
H0: μ1 = μ2
z = (xbar1-xbar2) / sqrt( σ1^2 /n1 +σ2^2/n2)
sqrt( σ1^2 /n1 +σ2^2/n2) = sqrt ( 4.5^2/30 + 4^2/30) = 1.099242
z = (67.5 - 64.6) = 2.9 / 1.099242
z =2.6382
b)
Find the p-value
P( z > 2.6382) = 0.0041
Since the p-value is smaller than the significance level of 0.05, we reject the null hypothesis; there is sufficient evidence to conclude that the average retirement age at ShopRite is higher than it is in Boscov’s.
To solve this problem, you need to perform a hypothesis test and calculate the p-value.
a) Hypothesis testing involves comparing the sample means and determining if there is enough evidence to support a claim about the population mean. In this case, we want to test if the average retirement age at ShopRite is higher than at Boscov's.
The null hypothesis (H0) assumes there is no difference between the two means:
H0: μ1 = μ2
The alternative hypothesis (Ha) assumes there is a difference:
Ha: μ1 < μ2
To perform the hypothesis test, you will use the t-test because the population standard deviations are unknown.
Calculate the test statistic t:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))
- x1 and x2 are the sample means (64.6 and 67.5)
- s1 and s2 are the sample standard deviations (4 and 4.5)
- n1 and n2 are the sample sizes (30 and 30)
t = (64.6 - 67.5) / sqrt((4^2 / 30) + (4.5^2 / 30))
Once you calculate the t-statistic, you can compare it to the critical value from the t-distribution. The critical value depends on the significance level (alpha) and the degrees of freedom (df = n1 + n2 - 2).
b) The p-value represents the probability of observing a test statistic as extreme as the one calculated (or more extreme) if the null hypothesis is true.
To determine the p-value, you will need to consult a t-table or use statistical software.
If the test statistic falls in the critical region (beyond the critical value), the p-value would be less than the significance level (alpha). In this case, alpha is 0.05.
Interpretation of results:
- If the p-value is less than 0.05, you can reject the null hypothesis and conclude that the average retirement age at ShopRite is higher than at Boscov's.
- If the p-value is greater than 0.05, you fail to reject the null hypothesis and cannot conclude that there is a significant difference in the average retirement age.
Remember, it's important to perform calculations accurately and consult appropriate resources, such as statistical tables or software, when needed.