Question

Two statistics teachers both believe that each has the smarter class. To put this to the test, they give the same final exam to their students. A summary of the class sizes, class means, and standard deviations is given below:n1 = 47, x-bar1 = 84.4, s1 = 18n2 = 50, x-bar2 = 82.9, s2 = 17Is there evidence, at an ?=0.04 level of significance, to conclude that there is a difference in the two classes? Carry out an appropriate hypothesis test, filling in the information requested.How do I find the standardized test statistic and the p-value?Your decision for the hypothesis test: A. Do Not Reject H0. B. Do Not Reject Ha. C. Reject H0. D. Reject Ha.

Answer 1

Answer:

We conclude that there is no difference between the two classes.

Step-by-step explanation:

We are given that two statistics teachers both believe that each has a smarter class.

A summary of the class sizes, class means, and standard deviations is given below:n1 = 47, x-bar1 = 84.4, s1 = 18n2 = 50, x-bar2 = 82.9, s2 = 17

Let $\mu_1$ = mean age of student cars.

 $\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1=\mu_2$      {means that there is no difference in the two classes}  

Alternate Hypothesis, $H_A$ : $\mu_1\neq \mu_2$      {means that there is a difference in the two classes}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

                         T.S.  =  $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$   ~  $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$  = sample mean age of faculty cars = 5.3 years  

$s_1$ = sample standard deviation of student cars = 3.6 years  

$s_2$ = sample standard deviation of student cars = 3.7 years  

$n_1$ = sample of student cars = 110  

$n_2$ = sample of faculty cars = 75  

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$  = $\sqrt{\frac{(47-1)\times 18^{2}+(50-1)\times 17^{2} }{47+50-2} }$  = 17.491

So, the test statistics =  $\frac{(84.4-82.9)-(0)}{17.491 \times \sqrt{\frac{1}{47}+\frac{1}{50} } }$  ~  $t_9_5$

                                     =  0.422    

The value of t-test statistics is 0.422.

Now, the P-value of the test statistics is given by;

P-value = P($t_9_5$ > 0.422) = From the t table it is clear that the P-value will lie somewhere between 40% and 30%.

Since the P-value of our test statistics is way more than the level of significance of 0.04, so we have insufficient evidence to reject our null hypothesis as our test statistics will not fall in the rejection region.

Therefore, we conclude that there is no difference between the two classes.