STA258 Lecture 24

• Some of the questions from previous finals.
• 1
  • a:
    • Youtuber MrBeast.
    • Challenge videos as well as philanthropy.
    • Is there a difference between the subscription rates between the two types of videos?
    • Group A: Challenge videos.
    • Group B: Philanthropy videos.
    • First 48 hours of the release.
    • Hypotheses:
    • $H_0:{\mu }_A={\mu }_B$
    • $H_a:{\mu }_A\ne {\mu }_B$
    • $\alpha =0.05$
    • Group A:
    • $\begin{array}{ccc}\text{Video}& \text{Views}& \text{New Subs}\\ 1& 1m& 300k\\ 2& 0.5m& 250k\\ 3& 300k& 270k\\ 4& 600k& 140k\\ 5& 1.2m& 400k\end{array}$
    • Group B:
    • $\begin{array}{ccc}\text{Video}& \text{Views}& \text{New Subs}\\ \\ 1& 1.2m& 450k\\ 2& 0.9m& 270k\\ 3& 700k& 220k\\ 4& 600k& 200k\\ 5& 900k& 250k\end{array}$
    • Values might not be right hard to see.
    • Since it's a sample, we need to calculate the sample means and standard deviations.
    • It also means we can't use $Z$ we need to use $t$ or something of that sort.
    • ${\mu }_A=\frac{1}{5}(300+250+270+140+400)=272k$
    • ${\mu }_B=\frac{1}{5}(450+270+220+200+250)=278k$
    • $s_A^2=\frac{1}{4}[(300-272{)}^2+(250-272{)}^2+(270-272{)}^2+(140-272{)}^2+(400-272{)}^2]=8770k$
      • $s_A=\sqrt{8770}=93.6482781475452k$
    • $s_B^2=\frac{1}{4}[(450-278{)}^2+(270-278{)}^2+(220-278{)}^2+(200-278{)}^2+(250-278{)}^2]=9970k$
      • $s_B=\sqrt{9970}=99.8498873309329k$
    • Suppose sample variances are equal, we can pool them together.
    • Formula: $s_p^2=\frac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}$
    • $s_p^2=\frac{4\cdot 8770+4\cdot 9970}{8}=9370k$
    • $s_p=\sqrt{9370}=96.7987603226405k$
    • $t=\frac{272-278}{96.7987603226405\sqrt{\frac{1}{5}+\frac{1}{5}}}=-0.0980057280577201$
    • $df=8$
    • $2P(T>\left|t\right|)$
    • $2P(T>|-0.09|)$
    • $2P(t_{(8;\frac{\alpha }{2})}>0.09)$
    • $2P(t_{(8;0.025)}>0.09)$
    • In R 2*pt(0.09, df=8, lower.tail=FALSE) gives $0.9304997$
    • $0.9304997>0.05$, so we fail to reject the null hypothesis.
    • • Solution
    • $H_0:p_A=p_B$
    • $H_a:p_A\ne p_B$
    • ${\hat{p}}_A=\frac{1360000}{4500000}=\frac{68}{225}=0.302222222222222$
    • ${\hat{p}}_B=\frac{1420000}{4500000}=\frac{71}{225}=0.315555555555556$
    • $\hat{p}=\frac{1360000+1420000}{(2)(4500000)}=\frac{139}{450}$
    • $Z=\frac{{\hat{p}}_A-{\hat{p}}_B}{\sqrt{\hat{p}(1-p)(n_1^{-1}+n_2^{-1})}}=-4.2$
    • $P(Z<-4.2)$ is very small, so we reject the null hypothesis.
    • Evidence suggests that the subscription rates are different between the two types of videos.
  • b:
    • Let $(x_1,y_1),\dots ,(x_n,y_n)$ be a set of $n$ data points.
    • Regression model: ${\hat{y}}_i={\hat{\beta }}_0+{\hat{\beta }}_1{x}_i^2$ for $i=1,\dots ,n$
    • ${\hat{y}}_i$ is the predicted value of $y_i$ based on the regression model.
    • Find the least squares estimates ${\hat{\beta }}_0$ and ${\hat{\beta }}_1$.
    • Consider using $u_i=x_i^2$.
    • ${\hat{y}}_i={\hat{\beta }}_0+{\hat{\beta }}_1{u}_i$
    • ${\hat{\beta }}_1=\frac{S_{XY}}{S_{XX}}$
      • $S_{XY}=\sum _{i=1}^n(u_i-\overline{u})(y_i-\overline{y})$
      • $S_{XX}=\sum _{i=1}^n(u_i-\overline{u}{)}^2$
      • $\overline{u}=\frac{1}{n}\sum _{i=1}^n{u}_i$
      • Group A:
      • $\begin{array}{ccc}\text{Video}& \text{Views}& \text{New Subs}\\ 1& 1m& 300k\\ 2& 0.5m& 250k\\ 3& 300k& 270k\\ 4& 600k& 140k\\ 5& 1.2m& 400k\end{array}$
      • Group B:
      • $\begin{array}{ccc}\text{Video}& \text{Views}& \text{New Subs}\\ \\ 1& 1.2m& 450k\\ 2& 0.9m& 270k\\ 3& 700k& 220k\\ 4& 600k& 200k\\ 5& 900k& 250k\end{array}$
        • ${\hat{\beta }}_1=\frac{\sum _{i=1}^n(u_i-\overline{u})(y_i-\overline{y})}{\sum _{i=1}^n(u_i-\overline{u}{)}^2}$
        • $\overline{u}=\frac{1}{n}\sum _{i=1}^n{u}_i$
        • $\overline{u}=\frac{1}{n}\sum _{i=1}^n{x}_i^2$
    • ${\hat{\beta }}_0=\overline{Y}-{\hat{\beta }}_1\overline{X}$
    • • Solution
    • $u_i=x_i^2$
    • ${\hat{y}}_i={\hat{\beta }}_0+{\hat{\beta }}_1{u}_i$
    • We want to minimize $f=\sum _{i=1}^n(y_i-{\hat{y}}_i{)}^2=\sum _{i=1}^n(y_i-{\hat{\beta }}_0-{\hat{\beta }}_1{u}_i{)}^2$
    • Differentiate with respect to ${\hat{\beta }}_0$ and ${\hat{\beta }}_1$ and set to 0.
      • $\frac{\partial f}{\partial {\hat{\beta }}_0}=-2\sum _{i=1}^n(y_i-{\hat{\beta }}_0-{\hat{\beta }}_1{u}_i)=0$
        • $\sum _{i=1}^n{y}_i{u}_i-{\hat{\beta }}_0\sum _{i=1}^n{u}_i-{\hat{\beta }}_1\sum _{i=1}^n{u}_i^2=0$
        • ${\hat{\beta }}_1=\frac{\sum _{i=1}^n{y}_i{u}_i-{\hat{\beta }}_0\sum _{i=1}^n{u}_i}{\sum _{i=1}^n{u}_i^2}$
        • ${\hat{\beta }}_1=\frac{\sum _{i=1}^n{y}_i{x}_i^2-{\hat{\beta }}_0\sum _{i=1}^n{x}_i^2}{\sum _{i=1}^n{x}_i^4}$
• 9
  • Analyse mrbeasts videos
  • $$\begin{array}{c}\text{Coefficients:}\\ & \text{Estimate}& \text{Std. Error}& \text{t value}& \text{Pr(>|t|)}\\ \text{(Intercept)}& -3.44& 0.097& -35& <2e-16\\ \text{log(view\_count)}& 0.99& 0.006& 173& <2e-16\end{array}$$
  • a:
    • Provide interpretations of both the intercept and slope of the best model for the data.
    • Are there any issues with the model?
    • ${\beta }_1=0.99$
    • For every log unit increase in view count, we expect the log like count to increase by 0.99 on average.
  • b:
    • Interpret the coefficient of determination for this model.
    • We have to look at the $R^2$ value for this model.
    • This tells us that $0.98$ of the uncertainty in the log like count can be explained by the log view count.
    • $R$ tells us that there is a strong positive linear relationship between log view count and log like count.
  • c:
    • Use the model to predict the number of likes for a ivdeo with $10,000,000$ views.
    • $\hat{y}=-3.465+0.991\log (10000000)=12.5080327900997$
    • $e^{12.5080327900997}=270501.464174755$
    • So we would expect a video with $10,000,000$ views to have approximately $270,501$ likes.