Question

In this exercise, you will investigate the relationship between growth and trade. The following table contains data on averagRound your response to Inice decimal places) The p-value is 0.088 (Round your response to three decimal places) Is the estima7 1 bwth and Tradeshare 2 Share Growth 3 0.140501976 1.918609 4 0.156622976 0.619799 5 0.157703221 4.308569 6 0.160405085 2.938 39 40 41 42 43 44 45 46 47 48 49 50 51 52 0.552722335| 1121508 0.560642541| 2.394881 0.56074965| 2.185974 0.567130983| 1.4

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Answer #1

The regression model is Growth on TradeShare.

The OLS regression result as estimated in excel is:

SUMMARY OUTPUT
Regression Statistics
Multiple R 0.211499042
R Square 0.044731845
Adjusted R Square 0.029324294
Standard Error 1.789139868
Observations 64
ANOVA
df SS MS F Significance F
Regression 1 9.293339116 9.293339116 2.903241733 0.093409233
Residual 62 198.463331 3.201021468
Total 63 207.7566701
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 0.961546419 0.5802784 1.657043272 0.102563148 -0.198413575 2.121506412 -0.198413575 2.121506412
Share 1.6820837 0.987201947 1.703890176 0.093409233 -0.291304874 3.655472274 -0.291304874 3.655472274

Thus, the regression equation is:

Predicted Growth = 0.961546418646467 + 1.68208369991324*Share

Null hypothesis: B1 = 0

Alternate Hypothesis : B1 <> 0

Conduct the t-test

The t-statistic as observed in the table above for the regression coefficient of Share is 1.70389017645379

Rounded to 3 decimal places, the t-statistic is 1.704

The actual p-value is 0.0934092325182143

At 10% level of significance, the critical p-value is 0.10

Since, the actual p-value is 0.0934092325182143 < 0.10, the null hypothesis is rejected.

Yes, the estimated coefficient is significant statistically.

Null Hypothesis is rejected and Alternate Hypothesis is accepted at 10% level of significance.

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