Question

Question1: It is known the population IQ score follows a normal distribution with mean as 100, SD as 10. A researcher is interested in studying if the average IQ of students from statistics courses on average has a higher IQ score than the population IQ score. To test this hypothesis, the researcher randomly collected a sample of 25 students from statistic class, the mean IQ score for this sample is 110. Compete for the hypothesis test at significant level.

Step 1: Hypothesis

H0:

H1:

Step 2: Criteria

Step 3: Calculation (hint: calculated the z score of )

Step 4: Conclusion (hint: if your obtained z score is more extreme than the critical z score, you could reject H0 and infer H1; if your obtained z score is less extreme than the critical z value, you will fail to reject H0)

Question2: If all information remains the same as question 1, except the researcher intended to test the hypothesis at a 1% significant level. What (i.e., H0; H1; ; one tail or two tail; Critical z; obtained z; reject or fail to reject) will be different?

Question3: If all other information remains the same as question 1, except the researcher’s hypothesis is “ The average IQ of students from statistics courses on average is different from IQ score than the population IQ score." What (i.e., H0; H1; ; one tail or two tail; Critical z; obtained z; reject or fail to reject) will be different?

Answer 1

The provided sample mean is Xˉ=110

and the known population standard deviation is σ=10,

and the sample size is n = 25.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ=100

Ha: μ>100

(2) Rejection Region

Based on the information provided, the significance level is α=0.05,

and the critical value for a right-tailed test is z_c = 1.64.

The rejection region for this right-tailed test is R={z:z>1.64}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z=5>zc​=1.64,

it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p = 0,

and since p=0<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 100, at the 0.05 significance level.

2)

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ=100

Ha: μ>100

(2) Rejection Region

Based on the information provided, the significance level is α=.01,

and the critical value for a right-tailed test is z_c = 2.33.

The rejection region for this right-tailed test is

R={z:z>2.33}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that

z=5>zc​=2.33,

it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p = 0,

and since p=0<.01,

it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 100, at the .01 significance level.

3)

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ=100

Ha: μ≠​100

(2) Rejection Region

Based on the information provided, the significance level isα=0.05,

and the critical value for a two-tailed test is z_c = 1.96.

The rejection region for this two-tailed test is

R={z:∣z∣>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that

∣z∣=5>zc​=1.96,

it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p = 0, and

since p=0<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 100, at the 0.05 significance

please like)