Question

Ten years ago the mean Math SAT score of all high school students who took the test in a small high school was 490, with a standard deviation of 80. This year, a researcher took the scores of a random sample of 16 students in the high school who took the SAT.
The mean score of these 16 students is (X bar) = 530. In addition, the researcher assumes that the population standard deviation continues to be σ = 80. The researcher will test if there is evidence that the scores in the district have changed with two approaches: (i) test of significance; and (ii) confidence interval.

1. Find H0 and Ha

2.z statistics

3.p value

4.What are your statistical conclusion and its interpretation? Use significance level, α = 0.05 (or 5%).

5.Find a 95% confidence interval for µ and interpret it. ( lower and upper bound)

6.Based on the hypotheses obtained in (a) and your 95% confidence interval for µ in (e), what is your conclusion? Is it the same as the conclusion of the test, i.e., (d)? Explain.

7.The researcher feels that the confidence interval for µ is too wide. So the researcher wonders how to increase of precision by decreasing the size of the confidence interval for µ. If the researcher can control only its sample size, what should be the researcher’s choice? How does it work? Explain.

Answer 1

The provided sample mean is \bar X = 530Xˉ=530 and the known population standard deviation is \sigma = 80σ=80, and the sample size is n = 16n=16.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ=490

Ha: μ≠​490

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Rejection Region

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the critical value for a two-tailed test is z_c = 1.96zc​=1.96.

The rejection region for this two-tailed test is R = \{z: |z| > 1.96\}R={z:∣z∣>1.96}

(2)Test Statistics

The z-statistic is computed as follows:

z=2

3.The p-value is p = 0.0455

(4) Decision about the null hypothesis

Since it is observed that ∣z∣=2>zc​=1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0455, and since p=0.0455<0.05, it is concluded that the null hypothesis is rejected.

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 490, at the 0.05 significance level.

5.

We need to construct the 95% confidence interval for the population mean \muμ. The following information is provided:

Sample Mean Xˉ =	530
Population Standard Deviation (σ) =	80
Sample Size (N)(N) =	16

The critical value for α=0.05 is The corresponding confidence interval is computed as shown below:

CI​=(490.801,569.199)​

lowerlimit=490.801

upper limit=569.199

INTERPRETATION:We are 95% confident that the true population mean μ is contained by the interval (490.801, 569.199)

6.population mean=490 which lies outside the confidence interval (490.801, 569.199).

Hence, we conclude that we reject the null hypothesis population mean is different than 490 which os same as we calculated above in the Z-test of single mean.

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