A deficiency of the trace element selenium in the diet can negatively impact growth, immunity, muscle and neuromuscular function, and fertility. The introduction of selenium supplements to dairy cows is justified when pastures have low selenium levels. Authors of a research paper supplied the following data on milk selenium concentration (mg/L) for a sample of cows given a selenium supplement (the treatment group) and a control sample given no supplement, both initially and after a 9-day period. Initial MeasurementTreatment Control11.3 9.19.9 8.710.1 9.78.5 10.810.4 10.910.9 10.611.8 10.19.8 12.310.6 8.810.4 10.410.2 10.911.3 10.49.2 11.610.9 10.910.88.2After 9 DaysTreatment Control138.3 9.4104 8.896.4 8.889 10.188 9.9103.8 8.9147.3 10.397.1 12.3172.6 9.4146.3 9.599 8.3122.3 8.7103 12.5117.8 9.1121.593(a)Use the given data for the treatment group to determine if there is sufficient evidence to conclude that the mean selenium concentration is greater after 9 days of the selenium supplement. (Use α = 0.05. Use a statistical computer package to calculate the P-value. Use μd = μinitial − μ9-day. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)t=df=P-value=State your conclusion. Reject H0. We have convincing evidence that the mean selenium concentration is greater after 9 days of the selenium supplement. Fail to reject H0. We have convincing evidence that the mean selenium concentration is greater after 9 days of the selenium supplement. Reject H0. We do not have convincing evidence that the mean selenium concentration is greater after 9 days of the selenium supplement. Fail to reject H0. We do not have convincing evidence that the mean selenium concentration is greater after 9 days of the selenium supplement.(b)Are the data for the cows in the control group (no selenium supplement) consistent with the hypothesis of no significant change in mean selenium concentration over the 9-day period? (Use α = 0.05. Use a statistical computer package to calculate the P-value. Use μd = μinitial − μ9-day. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)t=df=P-value=State your conclusion. Fail to reject H0. We have convincing evidence that the mean selenium concentration for cows in the control group changed significantly over the 9-day period. Reject H0. We have convincing evidence that the mean selenium concentration for cows in the control group changed significantly over the 9-day period. Fail to reject H0. We do not have convincing evidence that the mean selenium concentration for cows in the control group changed significantly over the 9-day period. Reject H0. We do not have convincing evidence that the mean selenium concentration for cows in the control group changed significantly over the 9-day period.(c)Would you use the paired t test to determine if there was a significant difference in the initial mean selenium concentration for the control group and the treatment group? Explain why or why not. Yes, the paired t test would be appropriate since the treatment and control groups were paired samples. No, the paired t test would not be appropriate since the treatment and control groups were not paired samples. No, the paired t test would not be appropriate because the number of sample differences is not large enough and the population distribution of differences is positively skewed. Yes, the paired t test would be appropriate because the sample differences can be viewed as a random sample from a population of differences, the treatment and control groups were paired samples, and we can assume the sample differences come from a normally distributed population of differences.