A statistics instructor is interested in the ability of students to assess the d

Question

A statistics instructor is interested in the ability of students to assess the difficulty of a test they have taken. This test was taken by a large group of students, and the average score was 85.8. A random sample of eight students was asked to predict this average score. Their predictions are shown below. Assuming a normal distribution, test the null hypothesis that the population mean prediction would be 85.8. Use a two-sided alternative and a 1% significance level. e 71 83 75 66 69 76 85 Click the icon to view the upper critical values of the Student's t distribution What are the null and alternative hypotheses for this test? O A. Ho: H285.8 O B. Ho: H 85.8 H1: H 85.8 H1: HK 85.8 O C. Ho: H 85.8 O D. Ho: S85.8 H1: H 85.8 H1: 85.8 For this test at the significance level o with sample mean x, hypothesized mean Ho, sample standard deviation s, and sample size n, what is the form of the decision rule? O A. Reje Ho if -tn -1 o/2 or reject Ho if o/2 O B. Reje Ho if tn 1,0 O c. Reje Ho if Ptn 1,0

Explanation / Answer

Given that,
population mean(u)=85.8
sample mean, x =75.25
standard deviation, s =6.5629
number (n)=8
null, Ho: =85.8
alternate, H1: !=85.8
level of significance, = 0.01
from standard normal table, two tailed t /2 =3.499
since our test is two-tailed
reject Ho, if to < -3.499 OR if to > 3.499
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =75.25-85.8/(6.5629/sqrt(8))
to =-4.547
| to | =4.547
critical value
the value of |t | with n-1 = 7 d.f is 3.499
we got |to| =4.547 & | t | =3.499
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -4.5468 ) = 0.0026
hence value of p0.01 > 0.0026,here we reject Ho
ANSWERS
---------------
null, Ho: =85.8
alternate, H1: !=85.8
test statistic: -4.547
critical value: -3.499 , 3.499
decision: reject Ho
p-value: 0.0026

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