Question

Please show steps in the calculation. Please make sure you include checking conditions for using the CLT.

Only Answer Question 3 and 4

The Human Resources (HR) Department of a certain college has asked all employees who were employed in 2018 to fill out a survey in December 2018. Three items on the survey were: “Your dental expense in 2018”, “Are you in a family with at least three other family members?”, and “Your medical expenses in 2018”. The manager of HR has randomly selected a sample of 169 surveys. Found that the sample average dental expense is $1600 per person with the sample standard deviation is $500. 70 of them are in a family of at least three other members. Also, the sample average medical expense is $2,450 per person and the sample standard deviation is $700.

1. The statistical inference will be made in parts (b), (c) and (d). Please clearly state the three parameters of interest.
2. Provide a 99% interval estimate of the average dental expenses per employee who was employed in the year 2018. please interpret your answer.
3. Using a 95% confidence interval, estimate the proportion of all employees who were employed in the year 2018 are in a family of at least three members. Interpret your answer.
4. As stated, the sample average medical expenses were $2,450. With your knowledge, the actual average price is more than $2,350. Obviously, 2,450 is more than 2,350 but this is only from a sample and is only $100 more, does it really mean that the actual average medical expenses are more than $2,350? Conduct a hypothesis test to prove the actual average medical expenses is more than $2,350 with a 5% level of significance. Use p-value to do a conclusion. Interpret your answer.

Answer 1

Provide a 99% interval estimate of the average dental expenses per employee who was employed in the year 2018.

We know the following from the sample

n=169 is the sample size

is the sample mean dental expense per person

is the sample standard deviation of dental expense per person

We estimate the population standard deviation using the sample

The standard error of mean is

Since the sample size is greater than 30, we can use CLT and use the normal distribution to approximate the distribution of sample mean.

The significance level for 99% confidence interval is

The right tail critical value is obtained using

Using the standard normal tables we get for z=2.58, P(Z<2.58)=0.995

Hence,

99% confidence interval is

ans: a 99% interval estimate of the average dental expenses per employee who was employed in the year 2018 is [150.77, 1699.23]

ans: there is a 99% chance that the true average dental expenses per employee who was employed in the year 2018 is between $150.77 and $1699.23

Using a 95% confidence interval, estimate the proportion of all employees who were employed in the year 2018 are in a family of at least three members. Interpret your answer.

We know the following from the sample

n=169 is the sample size

is the sample proportion of employees who were employed in the year 2018 are in a family of at least three members

The standard error of proportions is

Since both

are greater than 5, we can use CLT to use normal distribution as an approximate distribution of sample proportion

The significance level for 95% confidence interval is

The right tail critical value is obtained using

Using the standard normal tables we get for z=1.96, P(Z<1.96)=0.975

Hence,

95% confidence interval is

ans: a 95% interval estimate of the proportion of all employees who were employed in the year 2018 are in a family of at least three members is [0.34,0.49]

ans: there is a 95% chance that the true proportion of all employees who were employed in the year 2018 are in a family of at least three members is between 0.34 and 0.49

Conduct a hypothesis test to prove the actual average medical expenses is more than $2,350 with a 5% level of significance

Let be the actual average medical expenses. We want to conduct a hypothesis test to prove the actual average medical expenses is more than $2,350, that is we want to test if

The hypotheses are

We know the following from the sample

n=169 is the sample size

is the sample mean medical expense per person

is the sample standard deviation of medical expense per person

We estimate the population standard deviation using the sample

The standard error of mean is

The hypothesized value of average medical expenses is

Since the sample size is greater than 30, we can use CLT and use the normal distribution to approximate the distribution of sample mean.

The test statistics is

This is a right tail test (The alternative hypothesis has ">")

The p-value is

We will reject the null hypothesis if the p-value is less than the significance level.

Here, the p-value is 0.0314 and it is less than the level of significance 0.05.

Hence we reject the null hypothesis

We conclude that, at 5% level of significance, there is sufficient evidence to support the claim that the actual average medical expenses is more than $2,350

Or, simply

The actual average medical expenses is more than $2,350.