Question

A paint manufacturer wishes to compare the drying times of two different types of paint. independent random samples of 11 cans of type a and 9 cans of type b were selected and applied to similar surfaces. the drying times (in hours) were recorded. the summary statistics are as follows. type a type b mean 76.8 hours 66.1 hours standard deviation 4.5 hours 5.1 hours sample size 11 9 construct a 95% confidence interval for the difference between the mean drying time for all type a paint and type b paint. the degree of freedom that is used for the construction of this 95% confidence interval is

Answer 1

Size of the first sample n_1=11

Size of the second sample n_2=9

Mean of the first sample \bar{x_1}=76.8

Mean of the second sample \bar{x_2}=66.1

Standard deviation of the first sample s_1=4.5

Standard deviation of the second sample s_2=5.1

Null Hypotheis: \mu_1=\mu_2 [mean drying time is same for two types]

Alternate Hypothesis:
`\mu_1\\eq \mu_2`

The test statistics:

Using the "Test for equality of two means"

`z=\frac{\bar{x_1}-\bar{x_2}}{\sqrt{(s_1^2)/(n_1)+(s_2^2)/(n_2)}}`

}=4.91939 [/tex]

Thus the calculated value is |z|=4.9194

Table value at 5% level of significance is 1.96.

Here the table value 1.96 is less than the calculated value. Thus we rejected H_0 at 5% level.

The 95% confidence interval for
`\mu_1-\mu_2:`

`(\bar{x_1}-\bar{x_2})\pm1.96[SE\,\, \,of \,\,\,((\bar{x_1}-\bar[x_2})`

`(76.8-66.1)\pm1.96[\sqrt{(4.5^2)/(11)+(5.1^2)/(9)}]`

`= 10.7\pm1.96*2.175`

`=(10.7+1.96*2.175, 10.7-19.6*2.175)`

`=(14.963, 6.437)`