Question

A market research firm reported the mean annual earnings of all family practitioners in the

United States was $178,258. A random sample of 59 family practitioners in Los Angeles had

mean earnings of = $191,350 with a standard deviation of s=$42,897. Do the data provide

sufficient evidence to conclude that the mean salary for family practitioners in Los Angeles

differs from the national average? Use the α=0.05 level of significance and the P-value method.

(a) What type of value is being measured? (a mean, proportion or standard deviation?)

(b) Is the population standard deviation known or unknown?

(c) What distribution is used to find the P-value for this type of measurement?

(d) What assumptions must hold for this method to be used? Check the assumptions.

(e) State the appropriate null and alternate hypotheses:

(f) What type of hypothesis test is this? (left tail, right tail or two tail)

(g) Compute the P-value.

(h) Based on the P-value, do you reject

(i) Is there enough evidence to conclude that the mean salary for family practitioners in Los

Angeles differs from the national average?

Answer 1

a). type of variable is being measured = mean

b).the population standard deviation is unknown

[ because we are given (s) the sample sd]

c).t distribution is used to find the P-value for this type of measurement

[ because we know the sample sd(s).]

d).assumptions:-

1. this is a simple random sample

[ it is verified , because in question, it is given that a random sample of 59 students....]

2. the variable is approximately normally distributed.

[ as the sample size = 59, is large enough, we can say that the variable(mean earnings) is normally distributed...so, there is no significant outlier.]

e). hypothesis:-

where, is  the mean salary for family practitioners in Los Angeles.

f). this is a two- tailed hypothesis test.

the given data are:-

sample mean ()= 191350

sample sd (s) = 42897

sample size (n) = 59

level of significance () = 0.05

test statistic be:-

degrees of freedom = (n-1) = (59-1) = 58

g). p value = 0.0225

[ i am using ti 84 plus calculator .

steps:-

2ND vars select tcdf in lower type 2.344, in upper type 100, in df type 58 enter enter.

your 1 tailed p value = 0.0112627536 0.01126

the two tailed p value = (2*0.01126) = 0.02252 0.0225 ]

h).decision:-

p value = 0.0225 < 0.05

so, we reject the null hypothesis.

i).conclusion:-

there is enough evidence to conclude that the mean salary for family practitioners in Los Angeles differs from the national average at 0.05 level of significance.

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