Please use Excel to solve this problem; I ONLY WANT part c) AND d) Use this link

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Please use Excel to solve this problem; I ONLY WANT part c) AND d)

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Listed below are altitudes (thousands of feet) and outside air temperature (degrees Fahrenheit) recorded during a Delta Flight from New Orleans to Atlanta.

a) Use Excel to make a scatterplot with altitude on the horizontal axis and temperature on the vertical axis. Give the scatterplot an appropriate title. part a) solved on Excel

b)   Use EXCEL to compute r for the paired data. --part b solved; [r =0.996776806 ]

c)  What is the value of the coefficient of determination? What percentage of the total variation can be explained by the linear relation between the variables? Does this suggest that altitude does a good job of explaining temperature?

d) Follow the steps below to perform a hypothesis test to determine if the linear correlation is significantly different from 0.

i. State the null & alternate hypothesis

H0 =

H1 =

ii. Use the r value found from part b) [r =0.996776806 ] to compute the t test statistic.

iii. State the degrees of freedom.

iv. Find the P-value for determining whether the correlation is significantly different from 0?

Explanation / Answer

Part c

The coefficient of correlation is given as r = 0.9968. The coefficient of determination or the value of R-square is defined as below:

Coefficient of determination = R2 = 0.9968*0.9968 = 0.99361

99.36% of the total variation can be explained by the linear relation between the variables. This implies that the altitude does a good job of explaining temperature.

Part d

i

Null and alternative hypotheses are given as below:

Null hypothesis: H0: The linear correlation is zero.

Alternative hypothesis: Ha: The linear correlation is significantly different from zero.

H0: = 0 versus Ha: 0

This is a two tailed test.

ii

The test statistic formula is given as below:

Test statistic = t = r*sqrt [(n - 2)/(1 – r2)]

We are given r = 0.9968 and n = 7

iii

Degrees of freedom = n – 2 = 7 – 2 = 5

Test statistic = t = 0.9968*sqrt[(7 – 2)/(1 – 0.9968*0.9968)]

Test statistic = t = 27.884

iv

P-value = 0.000001 (by using t-table or excel)

Consider alpha = 0.01

P-value < alpha

At 1% level of significance, we reject the null hypothesis that the linear correlation is zero.

So, we conclude that the linear correlation is significantly different from zero.

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