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6. A psychology professor of a large class became curious as to whether the students who turned in tests first scored differently from the overall mean on the test. The overall mean score on the test was 75 with a standard deviation of 10; the scores were approximately normally distributed. The mean score for the first 20 students to turn in tests was 78. Using the .05 significance level, was the average test score earned by the first 20 students to turn in their tests significantly different from the overall mean?

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nuhulawal20

Answer: Z is less than Zc ∴ 1.342 < 1.96

Therefore, Null hypothesis is not Rejected.

There is no sufficient evidence to claim that students turning in their test first score is significantly different from the mean.

Step-by-step explanation:

Given that;

U = 75

X = 78

standard deviation α = 10

sample size n = 20

population is normally distributed

PROBLEM is to test

H₀ : U = 75

H₁ : U ≠ 75

TEST STATISTIC

since we know the standard deviation

Z =  (X - U) / ( α /√n)

Z = ( 78 - 75 ) / ( 10 / √20)

Z = 1.3416 ≈ 1.342

Now suppose we need to test at ∝ = 0.05 level of significance,

Then Rejection region for the two tailed test is Zc = 1.96

∴ Reject H₀ if Z > Zc

and we know that Z is less than Zc ∴ 1.342 < 1.96

Therefore, Null hypothesis is not Rejected.

There is no sufficient evidence to claim that students turning in their test first score is significantly different from the mean.

Testing the hypothesis, it is found that since the p-value of the test is 0.1802 > 0.05, which means that the average test score earned by the first 20 students to turn in their tests was not significantly different from the overall mean.

At the null hypothesis, we test if the mean is of 75, that is:

[tex]H_0: \mu = 75[/tex]

At the alternative hypothesis, we test if the mean is different of 75, that is:

[tex]H_1: \mu \neq 75[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which:

  • X is the sample mean.
  • [tex]\mu[/tex] is the value tested at the null hypothesis.
  • [tex]\sigma[/tex] is the standard deviation.
  • n is the size of the sample.

For this problem, we have that:

[tex]X = 78, \mu = 75, \sigma = 10, n = 20[/tex]

The value of the test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{78 - 75}{\frac{10}{\sqrt{20}}}[/tex]

[tex]z = 1.34[/tex]

Since this is a two-tailed test, the p-value of the test is P(|z| < 1.34), which is 2 multiplied by the p-value of z = -1.34.

Looking at the z-table, z = -1.34 has a p-value of 0.0901.

2(0.0901) = 0.1802

The p-value of the test is 0.1802 > 0.05, which means that the average test score earned by the first 20 students to turn in their tests was not significantly different from the overall mean.

A similar problem is given at https://brainly.com/question/15535901

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