Question

3)-7) The mangers of a brokerage firm are interested in finding out if the number of new clients a broker brings into the firm affects the sales (in 1,000 $) generated by the broker. They sampled 25 brokers and determine the number of new clients they have enrolled in the last year and the data are partially presented in the table with the following results: y = 25.5+1.35X, SST=1430, SSR=1188.55, and -- X - )2-102. Broker Clients () Sales / 1 26 60 2 23 57 25 24 59 At 5% level of significance, which of the following represents the results of the relevant hypothesis test? no test no decision reject HO fail to reject HO QUESTION 6 3)-7) The mangers of a brokerage firm are interested in finding out if the number of new clients a broker brings into the firm affects the sales (in 1,000 $) generated by the broker. They sampled 25 brokers and determine the number of new clients they have enrolled in the last year and the data are partially presented in the table with the following results: y = 25.5+1.35X, SST-1730, SSR=1188.55, and --X. 2)2-102. Broker Clients (9) Sales 1 26 60 2 23 57 to 25 24 59 The 95% confidence interval estimate for the slope is 0.5480 to 2.1520 0.7650 to 1.9350 0.8002 to 1.8998 0.6864 to 2.0136

Answer 1

1. Here our hypothesis is given by

H0: The number of new clients a broker bring does not affect the sell.
H0: The number of new clients a broker bring affects the sell.

To test above hypothesis we use F test whose test statistic is given by

F = SSR/1 SSRes/(n - 2) F(1,n-2)

Here n=25 and SSR= 1188.55 and SST=1430; thus SSres= SST-SSR= 241.45, So

$F=\frac{1188.55/1}{241.45/(25-2)}=113.2187$

Now since the crtical value at alpha= 0.05 and df=(1,23) is $F_{critical}=3.408268$

Since the critical value is less than calculeted value thus the null hypothesis is rejected. 

2. Since

$t=\frac{\hat{\beta}_1-\beta_1}{se(\hat\beta_1)}\sim t_{n-2}$

where
B1
is slope and $\hat\beta_1$ is its estimates which is given as 1.35 and

$se(\beta_1)=\sqrt{\frac{SS_{Res}/n-2}{\sum(X-\bar{X})^2}}=\sqrt{\frac{241.45/23}{102}}=0.3208113$

Using t distriution property we can write

$P\left[-t_{\alpha/2,(n-2)}<\frac{\hat{\beta}_1-\beta_1}{se(\hat\beta_1)}<t_{\alpha/2,(n-2)}\right]=1-\alpha$

or

$P\left[\hat{\beta}_1-t_{\alpha/2,(n-2)}{se(\hat\beta_1)}<{\beta_1}<\hat{\beta}_1+t_{\alpha/2,(n-2)}{se(\hat\beta_1)}\right]=1-\alpha$

or

$P\left[\hat{\beta}_1-t_{0.05/2,(n-2)}{se(\hat\beta_1)}<{\beta_1}<\hat{\beta}_1+t_{0.05/2,(n-2)}{se(\hat\beta_1)}\right]=0.95$

or

$P\left[0.6864<{\beta_1}<2.0136\right]=0.95$

So option 4 is correct.