Question

The national mean SAT score is 1435. The standard deviation for all SAT scores is 42.7. A random selection of 31 tests has a(n) mean score of 1425. Can it be concluded that the mean SAT score will be 1435 based on α-level of 0.03?

a) Based on reading the problem, how do we know we are working with a hypotheses test (H0andH1H0andH1)? This question has two possible answers.

b) Please select the correct normal distribution based on the correct answer for H1H1?

c) To find our Critical Value(s) (CV), will we use invNorm(), invT()?

  Select an answer invT() invNorm()

d) Please explain the reason for the correct answer for step c. Select an answer This is a hypotheses test for the difference of population proportions (p₁ - p₂). The population standard deviation is not given The population standard deviation is given (σ). This is a hypotheses test for population proportion (p). This is a hypotheses test for the difference of population means (μ₁ - μ₂).

Round to Two Decimal Places

CV1 ≈ CV2 ≈

f) What formula will we use for the test value?

g) Find the values of each component of our test value formula. Then, use that information to find the test value.

h) Please select the one that best represents where the test value will be located on the correct normal distribution curve.

i) Does the test value land on the Noncritical Region or Critical Region?

Select an answer Noncritical Region Critical Region

j) Based on our evidence above, what conclusion can we make?

Select an answer Support Not Support Reject Not Reject  ? H₀ H₁

k) Please fill in the blanks in the sentence (express αα - level as a decimal).

Answer 1

- The provided sample mean is X 1425 and the known population standard deviation is 42.7, and the sample size is n = 31. 0 = (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M = 1435 Ha: M # 1435 This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.03, and the critical value for a two-tailed test is 2c = 2.17. The rejection region for this two-tailed test is R= {z:[2] > 2.17} (3) Test Statistics The Z-statistic is computed as follows: X – μο 1425 – 1435 z = = = -1.304 o/vn 42.7/31

(4) Decision about the null hypothesis - Since it is observed that |z 1.304 < Zc = 2.17, it is then concluded that the null hypothesis is not rejected. = Using the P-value approach: The p-value is p = 0.1923, and since p 0.1923 > 0.03, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 1435, at the 0.03 significance level.

Z-Test Results: t-stats = -1.304, p-value =0.1923 0.40 0.35 0.30 0.25 - 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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