Auniversity wants to compare out-of-state applicants' mean sat math scores (μ1) to in-state applicants' mean sat math scores (μ2). the university looks at 35 in-state applicants and 35 out-of-state applicants. the mean sat math score for in-state applicants was 540, with a standard deviation of 20. the mean sat math score for out-of-state applicants was 555, with a standard deviation of 25. it is reasonable to assume the corresponding population standard deviations are equal. to calculate the confidence interval for the difference μ1 − μ2, what is the number of degrees of freedom of the appropriate probability distribution?

64

64.87

68

69

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

represent the sample mean 1  (in-state applicants)

represent the sample mean 2  (out-of state applicants)

n1=35 represent the sample 1 size  (in-state applicants)

n2=35 represent the sample 2 size  (out-of state applicants)

sample standard deviation for sample 1  (in-state applicants)

sample standard deviation for sample 2  (out-of state applicants)

parameter of interest.

We are assuming that the population standard deviations are equal.

The confidence interval for the difference of means is given by the following formula:  

(1)

Where

The point of estimate for is just given by:

In order to calculate the critical value we need to find first the degrees of freedom, given by:  

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐