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# Confidence Interval

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Please provide step by step instructions using actual numbers instead of formulas for solving the following 3 problems, thanks

3. Determine whether the average number of calories in a homemade cookie is more than a store-bought one. Estimate the difference in the mean calories between the two types using a 90% confidence interval.
storebought n=45 x-bar=179 s=4

4. A childcare agency was interestin in examining the amount that families pay per child per month for childcare outside the home. A randome sample of 64 families was selected and the mean and the standard deviation were computed to be \$675 and \$80 respectively.
a. Find a 95% confidence interval for the true average amount spent per child per month for childcare outside the home.
b. Interpret the interval found in part a.
c. A social worker claims that the average amount spent per child per month outside the home is \$700. Based on the interval in part a can this claim be rejected?
d. Find a 95% upper confidence bound for the true average amount spent per child per month on childcare outside the home.

##### Solution Summary

The expert determines whether the average number of calories in a homemade cookie is more than a store-bought one. The difference in the mean calories between the two types using a 90% confidence interval is estimated.

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3. Solution. Xbar=180, sx=2,n1=40, Ybar=179, sy=4, n2=45, 1-alpha=0.1, so a=alpha/2=0.05
t_0.05(n1+n2-2)=1.658.
therefore, we get the confidence interval of the difference in the mean of two types as ...

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###### Education
• BSc , Wuhan Univ. China
• MA, Shandong Univ.
###### Recent Feedback
• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
• "excellent work"
• "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
• "Thank you"
• "Thank you very much for your valuable time and assistance!"

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