Question

5. The servicing time at the drive-through lane of a fast food restaurant follows an exponential distribution. The average servicing time is 2 minutes. (4 points) YORKVILLE BUSI 1013 statistics for Business What is the probability that it takes more than 2.5 minutes to service a customer at the drive-through lane? b. a. What percent of customers at the drive-through lane will take between 1 to 3 minutes to service? Part B 1. A simple random sample of 50 customers is selected from an account receivable portfolio and the sample mean account balance is $1000. The population standard deviation ơ is known to be S200 (8 points) a Construct a 95% confidence interval for the mean account balance of the population. b. What is the margin of error for estimating the mean account balance at the 95% confidence level? Construct a 95% confidence interval for the mean account balance of the C. population if the sample mean account balance is obtained from a random sample of 100 instead of 50 d. Based on the results from a, and с., what can you say about the effect of a larger sample size on the length of the confidence interval at the same confidence level? 2. A simple random sample of 50 calls is monitored at an in-bound call center and the average length of the calls is 6 minutes. The population standard deviation σ is unknown. Instead the sample standard deviation s is also calculated from the sample and is found to be 4 minutes. (6 points) Construct a 99% confidence interval (using the t-distribution) for the average length of inbound calls b. What is the margin of error at the 95% confidence level? Do we need to make any assumption on the distribution of the length of in- bound calls at this call center? Why or why not? C. 3. According to a previous study, check out times at a supermarket are between 2 to 7 minutes (120 to 8400 seconds.) A survey involving a random sample of 3. According to a previous study, check out times at a supermarket are between 2 to 7 minutes (120 to 8400 seconds.) A survey involving a random sample of () YORKVILLE BUSI 1013 statistics for Business customers of the supermarket is planned and the first order of business is to decide on an appropriate sample size. The 95% level of confidence will be used. (4 points) a. What is the planning value for the population standard deviation to be used in the sample size formula for estimating the population mean check out time? b. How large should the sample size be if the desired margin of error is 30 seconds? 4. In a survey, the planning value for the population proportion is p 0.38. How large should the sample size for the survey be if the desired level of confidence is 95% and the desired margin of error is 0.0457(2 points)

Answer 1

5.

Let 'x' be our variable so

X ~ exp(λ =-)
  
Mean/Avg

P(X < x) =

AT

a. P(X > 2.5) = 1 - P(X < 2.5) =

е-0.5*2.5

Final Ans: 0.2865

b. P(1 < x < 3) = P(X<3) - P(X < 1)

-e-0.5*3-(1 -e-0.5*1) 1 =

=0.3834  

Final Ans: 38.34% customers at the drive through lne will take between 1 to 3 minutes to service.

Part B

1.

n = 50
= 1000 σ = 200

α = 5%

a. 95% Confidenc interval for mean $(\bar{x}-Z_{\alpha/2}*\frac{\sigma}{\sqrt{n}},\bar{x}+Z_{\alpha/2}*\frac{\sigma}{\sqrt{n}})$

Where

22.5% 1.96

200 ) 1.96 * V50 200 (1000-1.96 *-, 1000

Final Ans: (944.563, 1055.44)

b. Margin of error = z- score * Standard error where Standard error = $\sigma/\sqrt{n}$

Final Ans: 55.44

c. Will calculated similarly as 'a' but with n = 100

$(1000-1.96*\frac{200}{\sqrt{100}},1000+1.96*\frac{200}{\sqrt{100}})$

Final Ans: (960.8, 1039.2)

d. When the sample size is increased we get a better and a more accurate result than a smaller sample size. As we can that width in 'a' is 110.977 (1055.44 - 944.563) is greater than 78.4. That it is more distorted at n= 50 than at n = 100.

2.

$\bar{x}=5\;S_{x}=4\;n=50$$\alpha=1\%$

99% confidence interval for mean ($(\bar{x}-t_{\alpha/2,n-1}*\frac{S_{x}}{\sqrt{n}},\bar{x}+t_{\alpha/2,n-1}*\frac{S_{x}}{\sqrt{n}})$

Where $t_{0.5\%,49}=2.6806$ we subsitute the values

Final Ans: (3.484, 6.516)

b. Calculated similarly as 1.(b.)

MOE = t - score * S.E.

But $\alpha =5\%$ there t-score = 2.0102

Final Ans: 1.137

c. We need to assume that the sample comes from a normally distributed population, is random and independent. This is because when sample is large (n>30), we can apply the central limit theorem stating that as sample size increases the sample will follow a normal distribution.

3.

a. Planning value for population S.D. is range/4

Range = 7 - 2 = 5minutes (300 seconds)

Final Ans: 300/4 = 75 seconds (1.25 minutes)

This is calculated with the help of thumb rule of range (max - min) and S.D. This is a rough estimate.

There seems to be an error: 7 minutes is written as 8400 seconds. Please view the question.

b. Using S.D. calculated in 'a' as our population S.D. We have

MOE = z-score*S.E.

z-score at 95% = 1.96 MOE = 30 sec Therefore our equation will be

75 30 - 1.96.

Final Ans: n = 24.01 $\approx$ 25 customers

4.

MOE = 0.045 p* = 0.38

For population proportion

MOE = z-score * $\sqrt{\frac{p^{*}(1-p^{*})}{n}}$

z-score for 95% = 1.96

$0.045=\sqrt{\frac{0.38(1-0.38)}{n}}$

Final Ans: 446.95$\approx$447