Let's correct the calculation for the standardized test statistic (t) in step iv. ### iv) Standardized Test StatisticThe standardized test statistic (t) is calculated as: \[ t = \frac{\bar{D} - \mu_D}{s_D / \sqrt{n}} \] Under the null hypothesis, \( \mu_D = 0 \): \[ t = \frac{-2.17 - 0}{1.93 / \sqrt{6}} \] First, calculate the standard error: \[ \text{Standard Error} = \frac{s_D}{\sqrt{n}} = \frac{1.93}{\sqrt{6}} = \frac{1.93}{2.45} = 0.787 \] Now, calculate the t-value: \[ t = \frac{-2.17}{0.787} = -2.76 \] The correct value for the standardized test statistic is \( t = -2.76 \). ### v) P-value for the Standardized Test StatisticFor a two-tailed test with \( t = -2.76 \) and \( df = n-1 = 5 \): Using a t-distribution table or calculator, the p-value for \( t = -2.76 \) with 5 degrees of freedom is approximately 0.020. ### vi) ConclusionBased on the p-value (0.020), which is less than the typical significance level of 0.05, we reject the null hypothesis. This suggests that there is a statistically significant difference in the reading age between the experimental group and the control group. So, the correct standardized test statistic should be \( t = -2.76 \). Let's fill in the correct value for the standardized test statistic in the image: \[ t = -2.76 \] This should be entered in the box for the standardized test statistic.

Let's correct the calculation for the standardized test statistic (t) in step iv. ### iv) Standardized Test StatisticThe standardized test statistic (t) is calculated as: \[ t = \frac{\bar{D} - \mu_D}{s_D / \sqrt{n}} \] Under the null hypothesis, \( \mu_D = 0 \): \[ t = \frac{-2.17 - 0}{1.93 / \sqrt{6}} \] First, calculate the standard error: \[ \text{Standard Error} = \frac{s_D}{\sqrt{n}} = \frac{1.93}{\sqrt{6}} = \frac{1.93}{2.45} = 0.787 \] Now, calculate the t-value: \[ t = \frac{-2.17}{0.787} = -2.76 \] The correct value for the standardized test statistic is \( t = -2.76 \). ### v) P-value for the Standardized Test StatisticFor a two-tailed test with \( t = -2.76 \) and \( df = n-1 = 5 \): Using a t-distribution table or calculator, the p-value for \( t = -2.76 \) with 5 degrees of freedom is approximately 0.020. ### vi) ConclusionBased on the p-value (0.020), which is less than the typical significance level of 0.05, we reject the null hypothesis. This suggests that there is a statistically significant difference in the reading age between the experimental group and the control group. So, the correct standardized test statistic should be \( t = -2.76 \). Let's fill in the correct value for the standardized test statistic in the image: \[ t = -2.76 \] This should be entered in the box for the standardized test statistic.

What is the value of the test statistic?

(a) A statistician randomly sampled 100 observations and found X̄ = 106 and s = 35. Calculate the t statistic and p value for testing.     H0 : μ = 100 against HA : μ > 100.Carry out the test at the 1% level of significance.(b) Repeat part (a), with s = 25.(c) Repeat part (a), with s = 15.

tutor promises to improve GMAT scores of students by more than 36 points after three lessons. To see if this is true, the tutor takes a sample of 49 students’ test scores after and before they received tutoring. The mean difference was 39 points better after tutoring, with a standard deviation of the difference equal to 14 points. Let µD denote the mean of the difference: score after tutoring minus score before tutoring. Which of the following is the correct value of the test statistic?Multiple Choicez = 1.5000t48 = 1.5000t48 = 1.4643z = 1.4643

(a) A statistician randomly sampled 100 observations and found X̄ = 106 and s = 35. Calculate the t statistic and p value for testing.     H0 : μ = 100 against HA : μ > 100.Carry out the test at the 1% level of significance.(b) Repeat part (a), with s = 25.(c) Repeat part (a), with s = 15.Repeat question 1 using HA : μ ≠ 100.