Respuesta :
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: [tex] \mu_1\neq \mu_2 [/tex]
The test statistics:
Using the "Test for equality of two means"
[tex] z=\frac{\bar{x_1}-\bar{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} [/tex]
[tex] z=\frac{76.8-66.1}{\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9[tex] (76.8-66.1)\pm1.96(\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}) [/tex]}=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 [tex] \mu_1-\mu_2: [/tex]
[tex] (\bar{x_1}-\bar{x_2})\pm1.96[SE\,\, \,of \,\,\,((\bar{x_1}-\bar[x_2}) [/tex]
[tex] (76.8-66.1)\pm1.96[\sqrt{\frac{4.5^2}{11}+\frac{5.1^2}{9}}] [/tex]
[tex] = 10.7\pm1.96*2.175 [/tex]
[tex] =(10.7+1.96*2.175, 10.7-19.6*2.175) [/tex]
[tex] =(14.963, 6.437) [/tex]
