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Suppose that a lending rate influences profitability of commercial banks in Uganda. Use the regression below to answer the questions that follow: Profitability -44....

Question

Suppose that a lending rate influences profitability of commercial banks in Uganda. Use the regression below to answer the questions that follow: Profitability -44.73 + 0.7Lendingrate- R square = 0.81 SE(Bo) = 0.97 SE(8;) = 0.58, SE denotes standard error State the hypotheses (4mks) Conduct the t-test (4mks) Test whether not the estimated coefficient of ~Lending Rate" is statistically significant at 5% significance level (the critical value for the t- test at 5% significance level is 1.

Suppose that a lending rate influences profitability of commercial banks in Uganda. Use the regression below to answer the questions that follow: Profitability -44.73 + 0.7Lendingrate- R square = 0.81 SE(Bo) = 0.97 SE(8;) = 0.58, SE denotes standard error State the hypotheses (4mks) Conduct the t-test (4mks) Test whether not the estimated coefficient of ~Lending Rate" is statistically significant at 5% significance level (the critical value for the t- test at 5% significance level is 1.96) (4mks) Interpret the results (2mks) What does "R-squared = 0.81" mean? (2mks) lending rate is 18, what is the estimated profitability (Profitability is measured in USD)? (2mks)



Answers

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. Compute the $z$ value of the sample test statistic. (c) Find (or estimate) the $P$ -value. Sketch the sampling distribution and show the area corresponding to the $P$ -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level $\alpha ?$ (e) Interpret your conclusion in the context of the application.
Let $x$ be a random variable representing dividend yield of Australian bank stocks. We may assume that $x$ has a normal distribution with $\sigma=2.4 \% .$ A random sample of 10 Australian bank stocks gave the following yields. The sample mean is $\bar{x}=5.38 \% .$ For the entire Australian stock market, the mean dividend yield is $\mu=4.7 \%$ (Reference: Forbes). Do these data indicate that the dividend yield of all Australian bank stocks is higher than $4.7 \% ?$ Use $\alpha=0.01$

Part one. Okay, we at new sub to hat to the original equation. An estimated by old L s. The coefficient on new to hat is minus point Oh, 57 with 80 statistic of minus one point. Oh, eight. So what's the difference in the estimates of the return to education is large. It is not statistically significant. Part two. We now add near C two as an ivy along with near C four. The two stage least square estimate of beta one is now 0.157 with a standard error of point. Oh, 53 Okay, so the estimate is even larger. Yeah. Part three. Let you said I had to be the two stage least square residuals. Mm. We regret these on all exogenous variables, including near C two and near C four. The Choice square statistic, which is calculated by ticking and the number of observation multiply with our square from the regression is the product of 3000 and 10 observations and 0.4 which is the our square of the regression. You get 1.2. Yeah. The P value of this called square sta district is mhm 0.5 0.55 Yeah, so the over identifying restriction is not rejected. Mhm. Yeah,

18, which is big, each nodes new one equal to you too. On alternative five bus you want is not equal to a minute. Which will be the critical, very is good are the value in table four Correspondent toe probability, Open toe five 4.975 or 1.975 So is that different? To process or negative 1.96 So the rejection read, um, contain all values smaller than negative 1.96 and old very explosions in 1.96 So the testes statistic is that X one war minus X to war minus zero over squared off Simoni square over in one, once sigma squared over in tow, which, approximately equal to a negative, went to night. So if the value off the test statistic is within the rejection reasons that the knife was rejected so and negative 2.29 is smaller than negative 1.96 So we reject don't know hyper. So there is sufficient evidence so dejected

So in this video we will be making a scatter diagram then adjusting the axes and discussing what happens to the line of fit. I believe this is best served on dynamic software. So I've chosen to go to dez most dot com and I've already adjusted my scale to make sure that my grid is, my axes are in the same increments. So I come over here to the wrench and I already selected negative 1 to 10. I type them in negative 1 to 10. So that now my scales are the same now to graph the plot. I'm going to come over here to the plus sign on the left hand side, select table. And now I'm going to type in my values X values 1st +12345 and six and then my Y values 14 63 six seven. That's already plotted for me, which is fantastic. Now I'm asked to draw a line of fit. We're going to use technology to do that. So on this one in order to get my computer to find my line a fit, I type Y and then I need to underscore so that shift subtraction or underlying sign one. So I've done a subscript, then I move it over and I need to do the till day which is shift and then it's the little squiggly mark next to the one on the left hand side. M for my slope X shift minus sign or underlying one plus sign be. So this tells my computer the computer program to use these values from my grid. Now it's already done the line of best fit according to how it's been pre programmed and I'm told that the slope is 0.942857 So here's my line if it so here's my scatter plot. Notice this value right here. This is the value 1, 1/1 up one. This was 24 4336 It looks like I can't quite get that point right there because it's too close to the line. But those are my other two values. So now what I need to do is I need to increase the y axis, here's the X axis, here's the y axis. I need to increase the y axis so that it is doubled. So I'm gonna come over here to my wrench and I'm going to change my Y axis and simply multiply the values by two. And by doing that you can see with my Y axis, double the length of my X. Axis, my line if it appears to be flatter. However the slope remains the same. Now what I need to do is I need to take my Y axis and I need to make it half my X. Axis. And so first thing I'm gonna do is change it back to normal. So here we are where we were at the beginning where it's the same, click on the wrench again. But instead of taking the y axis and dividing in half, I'm actually going to double the X axis because that way it stays all on my screen so I'm going to double the X axis. So now my Y axis is actually half the scale. See it goes to 10 of my X axis and you can see here that the line if it appears to be steeper. However, the slope remains the same. So in this particular situation with our scattered diagram, it was actually pretty straightforward in that when I changed my y axis, my value that the the appearance of the steepness of the slope of my life seem to change. So how do these slopes compare? Well, here's how I would say it, and I'm actually when the scale is changed, when the scale of the Y axis is changed, the slope of the line of fit appears to change. However, it actually stays the same A Y axis scale that is twice the X axis appears flatter. You might want to pause the video to copy or write this down, and then the Y axis scale, that is half the X axis scale appears stever.

Yeah. Part one. This is what we get when we estimate this regression equation using profit. There's only one explanatory variable and this variable white dummy Takes on two values. So there are two predicted values. The estimated probabilities of loan approval for white and non white applicants. The probability is .7084 White applicant and for non white applicant. Yeah, It is .908. These are the same to the fitted values from the linear probability model. This is always the case when the independent variables in a binary respond model are mutually exclusive. An exhaustive binary variables. The predictive probabilities would be the same whether we use it's a linear probability models, profit are logic models. Yeah. Intersect .708 is a proportion of loans approved. Okay, I just flip the case. This ratio is a proportion of loans approved four nonwhites. This one is a proportion of loans approved for white applicants. Yes. Part to you. The profit estimate I'm white is 0.5 20. With a standard error of point zero 97 It's there's still very strong evidence of discrimination against non whites to make these estimate comparable to the L p. M estimate in part three. Yeah, we can divide this estimate. Yeah. What? Yeah. Bye 2.5. And you can get .208 Compare with .1-9 in the L p. M. Model. Ems for model. So in the L P M. Yeah, part three. Yeah. What we use logic instead of profit. What we get for the estimate on white is yes .938. and the standard error is .173 part forward. In order to make profit. And logic estimates roughly comparable. We can multiply the logic estimates by .6-5. In this part, the scaled lodging coefficient is point 6 to 5. Yeah, times .938. Yeah. And you would get The scale logic coefficient to be .586, which is quite close to the property estimate. Yeah.


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