Question

SAT reading and writing section scores of a random sample of twenty 11th-grade students in a certain high school are given below. 380 520 480 510 560 630 670 490 500 550 400 350 440 490 620 660 700 730 740 560 Test if the standard deviation of the reading and writing section SAT score of the students in this school is higher than 100 using. What is the p-value (round off to second decimal place)?

Answer 1

Answer:

No, standard deviation of the reading and writing section SAT score of the students in this school is not higher than 100.

P - value of test = 16.8% .

Step-by-step explanation:

We are given the random sample of reading and writing section scores of twenty 11th-grade students in a certain high school ;

380, 520, 480, 510, 560, 630, 670, 490, 500, 550, 400, 350, 440, 490, 620, 660, 700, 730, 740, 560.

We have to test if the standard deviation of the reading and writing section SAT score of the students in this school is higher than 100 or not.

       Null Hypothesis, $H_0$ : $\sigma$ = 100

Alternate hypothesis, $H_1$ : $\sigma$ > 100

So, the test statistics we use here will be ;

                   $\frac{(n-1)s^{2} }{\sigma^{2} }$ follows $\chi^{2}_n__-_1$

 where, n = sample size = 20

              $\sigma$ = Population standard deviation

        Xbar = $\frac{\sum X_i}{n}$ = $\frac{Sum of each above score}{20}$ = 549

   s = Sample standard deviation = $\frac{\sum (X_i - Xbar)^{2}}{n-1}$ = 114.61

       Test Statistics = $\frac{(20-1)*114.61^{2} }{100^{2} }$ follows $\chi^{2}__1_9$

                               = 24.96

Now, we assume level of significance to be 5% so at this level chi-square % table gives critical value of 30.14 at 19 degree of freedom and since our test statistics falls below this value or is less than critical value as 24.96 < 30.14 so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that standard deviation of the reading and writing section SAT score of the students in this school is equal to 100.

To calculate P-value we know that;

           P($\chi^{2}__1_9$ > 24.96) = This shows that P-value will lie between 10% to 20%

                                         by using chi-square % table.

Solving we get that P -value is 0.167878 or 16.8% .