#4 It must be in the format of the second picture. **Transcript of the Educational Video on Statistical Analysis**---**Heading:** Statistical Analysis and Probability Calculation---**Random Variable Section:**- **Random Variable:** \(\overline{X}\)- **Special considerations/characteristics:** (None specified)---**Given Information:**- \(P(\overline{X} \leq 6820) = 0.0030\)**Question Being Addressed:**- Determine \(P\) for \(\overline{X} \leq 6820\).---**Sample Information:**- Sample Size (\(n\)): 50- Sample Mean (\(\overline{x}\)): 6820---**Sampling Distribution Analysis:**- **Type:** Normal Distribution- **Why Justified:** Central Limit Theorem (CLT) applies since \(n\) is sufficiently large.- **Test Statistic:** \(Z\)- **Test Statistic Value Calculation:** - Population Mean (\(\mu_x\)): 7500 - Population Standard Deviation (\(\sigma_x\)): 1750 - Standard Error (\(\sigma_{\overline{x}}\)): \(\frac{1750}{\sqrt{50}} = 247.4874\) - \(Z = \frac{\overline{x} - \mu_x}{\sigma_{\overline{x}}} = \frac{6820 - 7500}{247.4874} = -2.75\)---**Probability Determination (C Section):**- **Calculate:** - \(P(\overline{X} \leq 6820)\) - Equivalent to finding \(P(Z \leq -2.75)\) - Result: \(0.0030\)**Graphical Representation:**- **Normal Distribution Curve:** - The graph illustrates the normal distribution with a shaded region indicating the probability for \(Z \leq -2.75\). - The mean (center) of the distribution is marked, with the negative Z-score on the left tail showing the area of interest.---**Additional Materials:**- Showing usage of a TI calculator for calculation verification.---This educational material aids in understanding the application of the Central Limit Theorem in determining probabilities related to sample means using normal distribution. ### Statistical Analysis Problems---**Problem 4: Graduation Rate Analysis**Women athletes at a university have a long-term graduation rate of 67%. Over the past several years, a random sample of 38 women athletes at the school showed that 21 eventually graduated. Does this indicate that the population proportion of women athletes who graduate from this university has changed (either way)? Use a 1% level of significance.---**Problem 5: Fishing Efficiency Study**Is it better to fish from the shore or a boat? Let \( x \) be a random variable representing the number of hours taken to catch a fish. The table below contains the data taken over several months for \( B \) from the shore, and \( A \) from a boat.| | Oct | Nov | Dec | Jan | Feb | Mar | Apr ||---------|-----|-----|-----|-----|-----|-----|-----|| B (shore) | 1.6 | 1.8 | 2.0 | 3.2 | 3.9 | 3.6 | 3.3 || A (boat) | 1.5 | 1.4 | 1.6 | 2.2 | 3.3 | 3.0 | 3.8 |Use a 1% level of significance to determine if there is a difference in the population mean hours per fish using a boat compared with fishing from shore. (This is a paired difference problem.)---**Problem 6: Hay Fever Rate Comparison**In a location, two random samples were taken concerning the rate of hay fever per 1000 population. Sample 1 (\( n_1 = 14 \)) contained people under 25, sample 2 (\( n_2 = 16 \)) contained people over 50.\[\begin{align*}\overline{X}_1 &= 109.5, \, s_1 = 15.41 \\\overline{X}_2 &= 99.36, \, s_2 = 11.57 \\\end{align*}\]Assume that the hay fever rate in each group has an approximately normal distribution. Do the data indicate that the over 50 group has a significantly lower hay fever rate? (Use \( \alpha = 5\% \))(Note: Several of these problems have been

To testH0:p=67%=0.67H1:p≠0.67Given n=38 and x=21. Then,p^=xn=2138=0.5526Then,z=p^-p0p01-p0n=0.5526-0.670.671-0.6738=-0.11740.0058=-0.11740.07628=-1.5391≃-1.54So, the p-value isPZ<-1.54=0.0618 To draw the graph, simply replace -2.75 by -1.54 and the rest is same.Since, p-value>0.01, failed to reject H0.Therefore, the women athletes at a university do not have a long term gratuate rate of 67%.