Question

The California Health Survey involved 51,048 randomly selected adults in California. The author selected 100 of those subjects and obtained these results. Among the 53 females, the mean height is 63.7 in. and the standard deviation is 4.5 in.: among the 47 males, the mean height is 69.2 in. and the standard deviation is 3.1 in. Test the claim that the mean height of an adult California female is greater than the mean height of an adult California male when a = 0.05.

Answer 1

Answer:

As we have not any information of the population standard deviation so we will use the two sample t test

Null hypotheis: Mean of height of the female in california is less than or equal to the mean height of the male of the california

Alternate hypotheis: Mean of height of the female in california is greater than the mean height of the male of the california

Based on the two sampel t test

Variance = standard deviation^2

variance of female = 4.5*4.5 =25.25

variance of male = 3.1*3.1 =9.61

t = (mean height of the female - mean height of male)/ sqrt((variance of heigh of female/ no. of observation in female) +(variance of heigh of male/ no. of observation in male))

t = (63.7 - 69.2)/sqrt(25.25/53 + 9.61/47)

t = -5.5/0.8251

t = -6.6654

Now if we observe the critical t value at 95 % CI on right side at 99 degree of freedom is 1.66309

So calculate t value is less than critical t value so we fail to reject the null hypothesis and states that Mean of height of the female in california is less than or equal to the mean height of the male of the california.

So we can reject the claim as we don't have significantly evidence that average female height is higher than average male height in california.