## Understanding P-values and Hypothesis Testing### P-value Ranges:- \(0.050 < \text{P-value} < 0.100\)- \(0.010 < \text{P-value} < 0.050\)- \(\text{P-value} < 0.010\)### Graphical Representation of Sampling DistributionsFour graphs are displayed, each illustrating a normal distribution curve with shaded areas representing different P-value regions. These visual aids help in understanding the significance of the P-value concerning the null hypothesis.1. **Graph 1**: - A standard normal distribution curve (bell-shaped). - The right tail is shaded, showing the area corresponding to the P-value.2. **Graph 2**: - Similar to Graph 1. - The shaded region is larger, indicating a larger P-value area towards the positive and negative tails.3. **Graph 3**: - A normal curve with a significant portion of both tails shaded, implying a small P-value.4. **Graph 4**: - The curve shows a smaller area shaded beyond the critical value, indicating a highly significant P-value.### Questions and Decision Making**(d) Evaluate Hypothesis Testing Decisions:**Choose based on whether the null hypothesis should be rejected:- At the \(\alpha = 0.01\) level, **fail to reject** or **reject** based on the data's significance: - **Fail to reject** the null hypothesis and conclude the data are not statistically significant. - **Reject** the null hypothesis and conclude the data are statistically significant.**(e) Conclusion in Application Context:**Interpret the outcome concerning boat and shore fishing:- **Fail to reject**: Insufficient evidence to claim a difference in mean hours between boat and shore fishing.- **Reject**: Sufficient evidence to claim a difference in mean hours between boat and shore fishing.---This example guides decision-making processes in hypothesis testing, emphasizing the practical implications of statistical outcomes. In this problem, assume that the distribution of differences is approximately normal. **Note**: For degrees of freedom \( d.f. \) not in the Student’s t table, use the closest \( d.f. \) that is smaller. In some situations, this choice of \( d.f. \) may increase the \( P \)-value by a small amount and therefore produce a slightly more “conservative” answer.Is fishing better from a boat or from the shore? Pyramid Lake is located on the Paiute Indian Reservation in Nevada. Presidents, movie stars, and people who just want to catch fish go to Pyramid Lake for really large cutthroat trout. Let row B represent hours per fish caught fishing from the shore, and let row A represent hours per fish caught using a boat. The following data are paired by month from October through April.| | Oct | Nov | Dec | Jan | Feb | Mar | Apr ||----|-----|-----|-----|-----|-----|-----|-----|| B: Shore | 1.4 | 1.8 | 2.0 | 3.2 | 3.9 | 3.6 | 3.3 || A: Boat | 1.4 | 1.3 | 1.6 | 2.2 | 3.3 | 3.0 | 3.8 |**Use a 1% level of significance to test if there is a difference in the population mean hours per fish caught using a boat compared with fishing from the shore.** (Let \( d = B - A \))### (a) What is the level of significance?Set the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?- \( H_0: \mu_d = 0; \ H_1: \mu_d \neq 0; \) two-tailed- \( H_0: \mu_d = 0; \ H_1: \mu_d < 0; \) left-tailed- \( H_0: \mu_d = 0; \ H_1: \mu_d > 0; \) right-tailed- \( H_0: \mu_d = 0; \ H_1: \mu_d \neq 0; \) two-tailed### (b) What

The paired data from Oct through April are given below:B:ShoreA: BoatOct1.41.4Nov1.81.3Dec21.6Jan3.22.2Feb3.93.3March3.63April3.33.8(a) The level of significance is The relevant null and alternative hypotheses are:Null Hypothesis i.e., there is no difference in the population mean hours per fish caught using a boat compared with fishing from the shore.Alternative Hypothesis i.e., there is a difference in the population mean hours per fish caught using a boat compared with fishing from the shore.This is a two-tailed test. Because the alternative Hypothesis includes 'not equal to' symbol.(b) Since the data are paired, a paired t-test is appropriate.The sampling distribution to be used is Student's t. We assume that the differences has an approximately normal distribution.The test statistic is given by the formula:where, is the sample mean of the differences. is the sample standard deviation of the differences.Calculations:B:ShoreA: Boatd=B-A1.41.401.81.30.521.60.43.22.213.93.30.63.630.63.33.8-0.52.6This implies:Thus, the value of the test statistic is 2.024.(c) The P-value is the probability of obtaining the test statistic value or a value even more higher assuming the null hypothesis is true.Degrees of freedom:Use Excel function "=T.DIST.2T(2.024,6)" to obtain the P-value. The result is shown below:Thus, the P-value can be expressed as .Since this is a two-tailed test, the area corresponding to the P-value is on both tails of the sampling distribution curve.Thus, Graph D is correct.Decision rule is:Reject if the P-value is less than or equal to significance level.Fail to reject if the P-value is greater than the significance level.Since the P-value is greater than 0.01, we fail to reject the null hypothesis and conclude that the data are not statistically significant.Option A is correct.Conclusion:Since the P-value is greater than 0.01, we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude a significant difference in the population mean hours per fish caught using a boat compared with fishing from the shore. Option A is correct.a) Level of significance is 0.01. versus b) The sampling distribution to be used is Student's t. We assume that the differences has an approximately normal distribution.The value of the test statistic is 2.024.c) The P-value can be expressed as .Graph D is correct.d) At the 0.01 significance level, we fail to reject the null hypothesis and conclude that the data are not statistically significant.Option A is correct.e) Fail to reject the null hypothesis. There is insufficient evidence to conclude a significant difference in the population mean hours per fish caught using a boat compared with fishing from the shore. Option A is correct.