The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of two English courses with a standard deviation of 0.7. The females took an average of three English courses with a standard deviation of 0.9. Are the means statistically the same? (Useα = 0.05)a) State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom. Round your answer to two decimal places.)b) What is the test statistic? (If using the z distribution round your answer to two decimal places, and if using the t distribution round your answer to three decimal places.)t=___________c) What is the p-value? (Round your answer to four decimal places.)

Answer:a) t distribution, and the degrees of freedom are given by [tex]df=n_m +n_f -2=29+16-2=43[/tex]  b) [tex]t=\frac{(2-3)-0}{\sqrt{\frac{0.7^2}{29}+\frac{0.9^2}{16}}}}=-3.848[/tex]c) [tex]p_v =2*P(t_{43}<-3.848)=0.0004[/tex]  Step-by-step explanation:Data given and notation  [tex]\bar X_{m}=2[/tex] represent the mean for the sample male[tex]\bar X_{f}=3[/tex] represent the mean for the sample female[tex]s_{m}=0.7[/tex] represent the sample standard deviation for the males [tex]s_{f}=0.9[/tex] represent the sample standard deviation for the females  [tex]n_{m}=29[/tex] sample size for the group male  [tex]n_{f}=16[/tex] sample size for the group female  t would represent the statistic (variable of interest)  Concepts and formulas to use  (a) State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom. Round your answer to two decimal places.)We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:  Null hypothesis:[tex]\mu_{m}-\mu_{f}=0[/tex]  Alternative hypothesis:[tex]\mu_{m} - \mu_{f}\neq 0[/tex]  We don't have the population standard deviation, so for this case is better apply a t test to compare means, and the statistic is given by:  [tex]t=\frac{(\bar X_{m}-\bar X_{f})-\Delta}{\sqrt{\frac{\sigma^2_{m}}{n_{m}}+\frac{\sigma^2_{f}}{n_{f}}}}[/tex] (1)And the degrees of freedom are given by [tex]df=n_m +n_f -2=29+16-2=43[/tex]  t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.b) What is the test statistic? (If using the z distribution round your answer to two decimal places, and if using the t distribution round your answer to three decimal places.)With the info given we can replace in formula (1) like this:  [tex]t=\frac{(2-3)-0}{\sqrt{\frac{0.7^2}{29}+\frac{0.9^2}{16}}}}=-3.848[/tex] c) What is the p-value? (Round your answer to four decimal places.) Since is a bilateral test the p value would be:  [tex]p_v =2*P(t_{43}<-3.848)=0.0004[/tex]  Comparing the p value with the significance level [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.