Question

The average American gets a haircut every 36 days. Is the average larger 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.

39, 32, 33, 38, 33, 38, 37, 31, 42, 47, 42, 45

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What can be concluded at the = 0.10 level of significance?

1. For this study, we should use Select an answer t-test for a population mean z-test for a population proportion
2. The null and alternative hypotheses would be:

H0:H0:  ? μ p  Select an answer = ≥ ≤ ≠      

H1:H1:  ? p μ  Select an answer < > ≠ =    

3. The test statistic ? z t  =  (please show your answer to 3 decimal places.)
4. The p-value =  (Please show your answer to 3 decimal places.)
5. The p-value is ? ≤ >  αα
6. Based on this, we should Select an answer fail to reject reject accept  the null hypothesis.
7. Thus, the final conclusion is that ... Select an answer At the 10% significance level, the data suggest there is sufficient evidence that μ higher than 36 At the 10% significance level, the data suggest there is insufficient evidence that μ higher than 36

Answer 1

From the given sample data : Sample size=n=12

Sample mean=$\bar{X}=\frac{\sum X}{n}=38.083$

Sample standard deviation=s=$\sqrt{\frac{\sum X^2-\frac{(\sum X)^2}{n}}{n-1}}=5.2129$

a. Here , the value of the population standard deviation is unknown.

Therefore , use t-distribution .

b. The null and alternative hypothesis is ,

$H_0:\mu=36;H_1:\mu>36$

c. The test statistic is ,

$t_{stat}=\frac{\bar{X}-\mu_0}{s/\sqrt{n}}=\frac{38.083-36}{5.2129/\sqrt{12}}=1.384$

d. Now , df=degrees of freedom=n-1=12-1=11

The p-value is ,

p-value=$P(t_{df}>|t_{stat}|)=P(t_{11}>1.384)=0.0969$ ; The Excel function is , =TDIST(1.384,11,1)

e. Decision : Here , p-value=0.0969 < 0.10 significance level.

Therefore , reject Ho.

Conclusion : At the 10% significance level, the data suggest there is sufficient evidence that μ higher than 36.