The amount of precipitation (in inches) in June of a recent year was measured in some randomly selected Michigan and Ohio cities (see below).
Assume that the mean amount of June precipitation in Michigan and Ohio cities are both approximately normally distributed.
Construct a 98% confidence interval for the difference of the mean amount of June precipitation in Michigan cities minus mean amount of June precipitation in Ohio cities.
Michigan Ohio
Lansing :3.46 Akron:3.15
BigRapids :3.27 Dayton:4.17
Monroe:3.62 Fremont:4.06
Marquette:2.68 Toledo:3.86
Alpena:2.68 Cincinnate:4.17

Respuesta :

Answer:

The 98% confidence interval for the difference of the mean amount of June precipitation in Michigan cities and Ohio cities is (-1.77, 0.29).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between two population mean is:

[tex]CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n-1)}\cdot\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}[/tex]

Compute the value of sample means and sample standard deviations from the provided data.

[tex]\bar x_{1}=3.142\\\\\bar x_{2}=3.882\\\\s_{1}=0.4396\\\\s_{2}=0.4283\\n_{1}=n_{2}=5[/tex]

The critical value of t for 98% confidence level and (n - 1) = 4 degrees of freedom is:

[tex]t_{\alpha/2, (n-1)}=t_{0.02/2, 4}=3.747[/tex]

Compute the 98% confidence interval for the difference of the mean amount of June precipitation in Michigan cities and Ohio cities as follows:

[tex]CI=(\bar x_{1}-\bar x_{2})\pm t_{\alpha/2, (n-1)}\cdot\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}[/tex]

     [tex]=(3.142-3.882)\pm3.747\cdot\sqrt{\frac{0.4396^{2}}{5}+\frac{0.4283^{2}}{5}}\\\\=-0.74\pm 1.0285\\\\=(-1.7685, 0.2885)\\\\\approx (-1.77, 0.29)[/tex]

Thus, the 98% confidence interval for the difference of the mean amount of June precipitation in Michigan cities and Ohio cities is (-1.77, 0.29).