Questions: The average American gets a haircut every 38 days. Is the average smaller for college students? The data below shows the results of a survey of 16 college students asking them how many days elapse between haircuts. Assume that the distribution of the population is normal. 43,44,30,38,35,26,27,39, 26,24,34,33,30,26,30,38 The final conclusion is a. The data suggest the population mean is significantly lower than 38 at a=0.01, so there is sufficient evidence to conclude that the population mean number of days between haircuts for college students is lower than 38. b. The data suggest the population mean number of days between haircuts for college students is not significantly lower than 38 at a=0.01, so there is insufficient evidence to conclude that the population mean number of days between haircuts for college students is lower than 38. c. The data suggest the population mean is not significantly lower than 38 at α= 0.01 , so there is sufficient evidence to conclude that the population mean number of days between haircuts for college students is lower than 38.

Solution Steps

To determine if the average number of days between haircuts for college students is significantly lower than 38 days, we can perform a one-sample t-test. The null hypothesis (H0) is that the mean number of days between haircuts for college students is 38 days. The alternative hypothesis (H1) is that the mean number of days is less than 38 days. We will use a significance level (alpha) of 0.01.

1. Calculate the sample mean and sample standard deviation of the given data.
2. Perform a one-sample t-test comparing the sample mean to the population mean of 38 days.
3. Determine the p-value from the t-test.
4. Compare the p-value to the significance level (0.01) to decide whether to reject the null hypothesis.

Given the data: \[ 43, 44, 30, 38, 35, 26, 27, 39, 26, 24, 34, 33, 30, 26, 30, 38 \]

The sample mean (\(\bar{x}\)) is calculated as: \[ \bar{x} = 32.6875 \]

The sample standard deviation (\(s\)) is: \[ s = 6.3321 \]

We perform a one-sample t-test to compare the sample mean to the population mean (\(\mu_0 = 38\)).

The t-statistic is calculated using: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]

Where \(n\) is the sample size (\(n = 16\)).

The p-value for the one-tailed test is: \[ p = 0.0022 \]

We compare the p-value to the significance level (\(\alpha = 0.01\)).

Since \( p < \alpha \) and the t-statistic is negative, we reject the null hypothesis.

Final Answer