The National Oceanic and Atmospheric Administration publishes temperature and precipitation information for cities around the world in Climates of the World. Data on average high temperature (in degrees Fahrenheit) in July and average precipitation (in inches) In July for \(48\) cities are on the Weiss Stats.

• For the estimations and predictions, use an average July temperature of \(83 \degree F\).
• For the correlation test, decide whether average high temperature in July and average precipitation in July are linearly correlated.

a. determine the sample regression equation.

b. find and interpret the standard error of the estimate.

c. decide at the \(5%\) significance level, whether the data provide sufficient evidence to conclude that the predictor variable is useful for predicting the response variable.

d. determine and interpret a point estimate for the conditional mean of the response variable corresponding to the specified value of the predictor variable.

e. find and interpret a \(95%\) confidence interval for the conditional mean of the response variable corresponding to the specified value of the predictor variable.

The value of a and b is given

\(a\approx 0.0067\)

\(b\approx 2.0481\)

We know that the regression line

\(\hat{y}=a+bx\)

Put the given value of a and b into the above equation

\(\hat{y}=a+bx\)

\(hat{y}=0.0067+2.0481x\)

The solution is \(hat{y}=0.0067+2.0481x\)

The value of a and b is given

\(a=79.4036\)

\(b= -0.3537\)

We know that the regression line

\(\hat{y}=a+bx\)

Put the given value of a and b into the above equation

\(\hat{y}=a+bx\)

\(hat{y}=0.0067+2.0481x\)

Calculate the standard error

\(s_{e}=\frac{\sigma}{\sqrt{n}}\)

After solving we will get

\(s_{e}=2.4134\)

The solution is \(s_{e}=2.4134\)

The predictor variable useful for predicting the response variable, when the slope of the sample regression line is nonzero.

\(H_{0}:\beta =0\)

\(H_{a}:\beta \neq 0\)

The p-value of the hypothesis will be

\(P\approx 0.8090\)

The null hypothesis is rejected.

\(P<0.05 \rightarrow\) Fail to Reject \(H_{0}\)

The null hypothesis is not rejected, so there is no sufficient evidence are provided for the sample.

Calculate the mean of the data of beef consumption this year

\(\bar{x}=\frac{\sum _{i-1}^{n}x_{i}}{n}\)

\(=\frac{70+58+...+13+40}{40}\)

\(=\frac{2069}{40}\)

\(\bar{x}=51.725\)

Calculate the mean of the data of beef consumption this year

\(\bar{x}=\frac{\sum _{i-1}^{n}x_{i}}{n}\)

\(=\frac{70+58+...+13+40}{40}\)

\(=\frac{2069}{40}\)

\(\bar{x}=51.725\)

Then calculate the variance

\(s^{2}=\frac{\sum (x-\bar{x})^{2}}{n-1}\)

\(=\frac{(70-51.725)^{2}+(58-51.725)^{2}+..+(40-51.725)^{2}}{40-1}\)

\(=\frac{14817.975}{39}\)

\(s^{2}=379.9481\)

Calculate the standard deviation

\(s=\sqrt{s^{2}}=\sqrt{379.9481}\approx 19.4923\)

Program:

Query:

• Sketch the graph of normal probability plot, boxplot, histogram and stem-and-leaf plot.