The Null and Alternative HypothesesLet’s say you are collecting data on employee age for your company in order to verify certain claims. For this, you collect data from a sample of 40 employees.Which of the following statements is correct?The null hypothesis can claim that the average age of employees is less than 38 years.The alternate hypothesis can claim that the average age of employees is not 32 years.The null hypothesis can claim that the average age of the sample of employees is 35 years.✕ IncorrectFeedback:Hypothesis testing is used to test claims about the population parameters (such as the population mean) using the sample data. Hence, the alternative hypothesis (or the null hypothesis, for that matter) cannot refer to a claim about the sample. Hence, this option is incorrect.The alternative hypothesis can claim that the average age of employees is greater than or equal to 27 years.

The null hypothesis always refers to a population statistic (e.g., the mean of X in the population).Select one:TrueFalse

Which of the following would be an appropriate null hypothesis?

In testing the hypothesesH0: μ = 75.HA: μ < 75,the p-value is found to be 0.042, and the sample mean is 80. Which of the following statements is true?Group of answer choicesNone of the above statements is correct.The probability of observing a sample mean at most as large as 75 from a population whose mean is 100 is 0.042.The probability that the population mean is smaller than 75 is 0.042.The smallest value of α that would lead to the rejection of the null hypothesis is 0.042.

Let's go through each part of the problem step by step. ### i) Null and Alternate HypothesesThe null hypothesis (H0) typically states that there is no effect or no difference. The alternate hypothesis (H1) states that there is an effect or a difference. - **Null Hypothesis (H0):** The mean difference in reading age between the experimental group and the control group is zero. (μD = 0) - **Alternate Hypothesis (H1):** The mean difference in reading age between the experimental group and the control group is not zero. (μD ≠ 0) ### ii) Sample Mean of DTo find the sample mean of D, sum all the differences and divide by the number of pairs. \[ D = [-3, -2, +1, -4, -4, -1] \] \[ \text{Sample Mean} (\bar{D}) = \frac{\sum D}{n} = \frac{(-3) + (-2) + 1 + (-4) + (-4) + (-1)}{6} = \frac{-13}{6} = -2.17 \] ### iii) Sample Standard Deviation of DTo find the sample standard deviation, use the formula: \[ s_D = \sqrt{\frac{\sum (D_i - \bar{D})^2}{n-1}} \] First, calculate each \( (D_i - \bar{D})^2 \): \[ (-3 + 2.17)^2 = 0.6889 \] \[ (-2 + 2.17)^2 = 0.0289 \] \[ (1 + 2.17)^2 = 9.8289 \] \[ (-4 + 2.17)^2 = 3.3769 \] \[ (-4 + 2.17)^2 = 3.3769 \] \[ (-1 + 2.17)^2 = 1.3689 \] Sum these values: \[ 0.6889 + 0.0289 + 9.8289 + 3.3769 + 3.3769 + 1.3689 = 18.6704 \] Then divide by \( n-1 \): \[ s_D = \sqrt{\frac{18.6704}{5}} = \sqrt{3.73408} = 1.93 \] ### 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, \( \mu_D = 0 \): \[ t = \frac{-2.17 - 0}{1.93 / \sqrt{6}} = \frac{-2.17}{0.787} = -2.76 \] ### v) P-value for the Standardized Test StatisticFor a one-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.

Which of the following represents the null and alternative hypotheses of the mean height of women is less  than 64