Using a 5% level of significance. Use 4 decimal places, unless the p-value is re

Question

Using a 5% level of significance. Use 4 decimal places, unless the p-value is really small (less than 0.0001), in which case use at least 2 significant figures.

Jeans Survey

Suppose that young adults own an average of 4 pairs of jeans. Survey 8 people from your class to determine if the average for students at De Anza College is higher than 4. DATA TO USE:   2, 3, 5, 5, 6, 6, 9, 15

a. Ho:
b. Ha:
c. In words, CLEARLY state what your random variable X or P' represents
d. State the distribution to use for the test.  
e. Test Statistic: t or z =
f. p-value =
In 1 – 2 complete sentences, explain what the p-value means for this problem.
g. Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the p-value.
h. Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using COMPLETE SENTENCES.
(i) alpha
(ii) decision
(iii) reason for decision
(iiii) Conclusion
i. Construct a 95% Confidence Interval for the true mean or proportion.   Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the Confidence Interval.
Confidence Interval: ( _______________ , ______________ )
j. Interpret the confidence interval.

Explanation / Answer

a. H0:mu<=4 (young adults own on an average atmost 4 pairs of jeans)

b. H1:mu>4 (yound adults own on an average more than 4 pairs of jeans)

c. The random variable X represent number of jeans owned by yound adults.

d. The level of measurement is nominal, and sample size is small with unknown population standard deviation. Therefore, use t distribution.

e. t=(xbar-mu)/(s/sqrt n), where, xbar is sample mean, mu is population mean, s is sample standard deviation, and n is sample size. The sample mean, xbar=sigma x/n, and sample standard deviation, s=sqrt[1/n-1 sigma (x-xbar)^2].

=(5.125-4)/(2.1/sqrt 8)

=1.52

f. p value at 7 df [df=n-1] is:0.087

Assuming the null hypothesis to be true, the probability of obtaining the observed t or more extreme result is 0.087.

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