9. According to a study, 16 – 26-year-olds make an average of 6.9 visits to shopping malls per week. A researcher wanted to check if this mean is true for the current population of 16 – 26-year old. A random sample of 45 such persons showed that the mean number of visits to shopping malls was 7.2 per week with a standard deviation of 1.3. Using the 2% significance level, is the mean number of visits to shopping malls per week for the current population of 16 – 26 – year – old is more than 6.9?
Answer:
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability.
To determine if the mean number of visits to shopping malls per week for the current population of 16-26-year-olds is more than 6.9, we can conduct a hypothesis test.
Step 1: State the hypotheses
The null hypothesis (H0) assumes that the mean number of visits is equal to 6.9:
H0: μ = 6.9
The alternative hypothesis (Ha) assumes that the mean number of visits is greater than 6.9:
Ha: μ > 6.9
Step 2: Set the significance level
The significance level, denoted as α, is set to 0.02 or 2% in this case.
Step 3: Calculate the test statistic
To perform the hypothesis test, we need to calculate the test statistic, which follows a t-distribution. The formula for the t-test is:
t = (x̄ - μ) / (s / √n)
Where:
x̄ = sample mean (7.2 visits)
μ = population mean (6.9 visits)
s = sample standard deviation (1.3)
n = sample size (45)
Plugging in the values, we get:
t = (7.2 - 6.9) / (1.3 / √45)
Step 4: Determine the critical value
The critical value is the value that separates the rejection region (where we reject the null hypothesis) from the acceptance region (where we fail to reject the null hypothesis). Since the alternative hypothesis is one-tailed (greater than), we need to find the critical value from the right side of the t-distribution table. With a 2% significance level and 44 degrees of freedom (n-1), we find the critical value to be approximately 2.433.
Step 5: Make the decision
If the test statistic t is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, let's calculate the t-value and compare it with the critical value.
Step 6: Calculate the t-value
t = (7.2 - 6.9) / (1.3 / √45)
t ≈ 0.716
Step 7: Compare the t-value with the critical value
Since the t-value (0.716) is less than the critical value (2.433), we fail to reject the null hypothesis.
Step 8: Make the conclusion
Based on the sample data, there is not enough evidence to conclude that the mean number of visits to shopping malls per week for the current population of 16-26-year-olds is more than 6.9 at the 2% significance level.