Question

the average american gets a haircut every 34 days. is the average different for college students? the data below shows the results of a survey of 12 college students asking them how many days elapse between haircuts. assume that the distribution of the population is normal.
21, 39, 41, 38, 25, 27, 30, 29, 25, 37, 36, 28
what can be concluded at the the \\( \alpha = 0.01 \\) level of significance level of significance?
a. for this study, we should use select an answer
b. the null and alternative hypotheses would be:
\\( h _ { 0 } : \\) ? select an answer
\\( h _ { 1 } : \\) ? select an answer
c. the test statistic ? = (please show your answer to 3 decimal places.)
d. the p - value = (please show your answer to 3 decimal places.)
e. the p - value is ? \\( \alpha \\)
f. based on this, we should select an answer the null hypothesis.
g. thus, the final conclusion is that ...
the data suggest the populaton mean is significantly different from 34 at \\( \alpha = 0.01 \\), so there is sufficient evidence to conclude that the population mean number of days between haircuts for college students is different from 34.
the data suggest the population mean number of days between haircuts for college students is not significantly different from 34 at \\( \alpha = 0.01 \\), so there is insufficient evidence to conclude that the population mean number of days between haircuts for college students is different from 34.
the data suggest the population mean is not significantly different from 34 at \\( \alpha = 0.01 \\), so there is sufficient evidence to conclude that the population mean number of days between haircuts for college students is equal to 34.

Explanation:

Step1: Determine the test type

Since the population standard deviation is unknown and the sample size \(n = 12\) (small sample, \(n<30\)) and the population is assumed to be normal, we use a \(t -\)test for a single mean.

Step2: State the null and alternative hypotheses

The null hypothesis \(H_{0}:\mu=34\) (the average number of days between haircuts for college students is the same as the average American). The alternative hypothesis \(H_{1}:\mu
eq34\) (the average number of days between haircuts for college students is different from the average American)

Step3: Calculate the sample mean \(\bar{x}\)

\(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\)
\(\sum_{i=1}^{12}x_{i}=21 + 39+41+38+25+27+30+29+25+37+36+28=376\)
\(\bar{x}=\frac{376}{12}\approx31.333\)

Step4: Calculate the sample standard deviation \(s\)

\(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(21 - 31.333)^{2}+(39 - 31.333)^{2}+(41 - 31.333)^{2}+(38 - 31.333)^{2}+(25 - 31.333)^{2}+(27 - 31.333)^{2}+(30 - 31.333)^{2}+(29 - 31.333)^{2}+(25 - 31.333)^{2}+(37 - 31.333)^{2}+(36 - 31.333)^{2}+(28 - 31.333)^{2}\)
\(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(- 10.333)^{2}+7.667^{2}+9.667^{2}+6.667^{2}+(-6.333)^{2}+(-4.333)^{2}+(-1.333)^{2}+(-2.333)^{2}+(-6.333)^{2}+5.667^{2}+4.667^{2}+(-3.333)^{2}\)
\(\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=106.777+58.777+93.444+44.444+40.111+18.777+1.777+5.444+40.111+32.111+21.777+11.111 = 474.777\)
\(s=\sqrt{\frac{474.777}{11}}\approx6.579\)

Step5: Calculate the \(t -\)test statistic

The formula for the \(t -\)test statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)
Substitute \(\bar{x}\approx31.333\), \(\mu = 34\), \(s\approx6.579\), \(n = 12\)
\(t=\frac{31.333-34}{6.579/\sqrt{12}}=\frac{- 2.667}{1.900}\approx - 1.404\)

Step6: Calculate the p - value

For a two - tailed \(t -\)test with \(n-1=11\) degrees of freedom. Using a \(t -\)distribution table or technology (e.g., TI - 84: tcdf(-100,-1.404,11)*2), the p - value is approximately \(0.182\)

Step7: Compare p - value and \(\alpha\)

Since \(\alpha=0.01\) and \(p - value=0.182>0.01\)

Step8: Make a decision about the null hypothesis

We fail to reject the null hypothesis.

Answer:

a. \(t -\)test for a single mean
b. \(H_{0}:\mu = 34\); \(H_{1}:\mu
eq34\)
c. \(t\approx - 1.404\)
d. \(p - value\approx0.182\)
e. \(p - value> \alpha\)
f. Fail to reject
g. The data suggest the population mean number of days between haircuts for college students is not significantly different from 34 at \(\alpha = 0.01\), so there is insufficient evidence to conclude that the population mean number of days between haircuts for college students is different from 34.