To test the hypothesis H0:µ^2 = 81 vs Ha:µ^2 ≠ 81, we can use the test statistic for a sample variance:
t = [(n-1)s^2] / σ^2
where n is the sample size, s^2 is the sample variance, and σ^2 is the hypothesized population variance.
Given that we have a sample size of 51, an average IQ of 108, and a standard deviation of 10, we can calculate the sample variance:
s^2 = (10^2) / (51-1)
s^2 = 100 / 50
s^2 = 2
The hypothesized population variance is σ^2 = 81.
Plugging in the values into the test statistic formula, we get:
t = [(51-1)(2)] / 81
t = 50(2) / 81
t = 100 / 81
t ≈ 1.23
To determine the outcome of the test, we compare the test statistic to the critical value(s) for a 5% significance level. Since the alternative hypothesis is not specific about the direction of the difference, we use a two-tailed test.
Looking up the critical t-value for a two-tailed test with a 5% significance level and 50 degrees of freedom (51-1), we find the critical t-value to be approximately ±2.01.
Since the test statistic (1.23) is less than the critical t-value (-2.01), we do not have enough evidence to reject the null hypothesis. Therefore, we fail to reject H0:µ^2 = 81 at the 5% significance level.
The IQ’s of 51 students are tested and it is found that the average IQ is 108 with a standard deviation of 10. Test H0:µ^2 = 81 vs Ha:µ^2 not =81 using a 5% significance level. Calculate the test statistic and outcome of the test
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