You may need to use the appropriate appendix table or technology to answer this question,
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A
sample of 67 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation
was 5.2 .
Provide a confidence interval for the population mean number of weekly customer contacts for the sales personnel,
(Round your answers to two decimal places.)
Provide confidence interval for the population mean number of weekly customer contacts for the sales personnel.
(Round your answers to two decimal places.)
contacts/week to

Question

You may need to use the appropriate appendix table or technology to answer this question,
Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A
sample of 67 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation
was 5.2 .
Provide a

confidence interval for the population mean number of weekly customer contacts for the sales personnel,
(Round your answers to two decimal places.)
Provide

confidence interval for the population mean number of weekly customer contacts for the sales personnel.
(Round your answers to two decimal places.)
contacts/week to

UpStudy AI · Accepted Answer

**Step 1. Compute the standard error (SE)**  
The standard error is given by  
$$
SE = \frac{s}{\sqrt{n}} = \frac{5.2}{\sqrt{67}}
$$  
Calculating the denominator,  
$$
\sqrt{67} \approx 8.19
$$  
Thus,  
$$
SE \approx \frac{5.2}{8.19} \approx 0.635
$$

**Step 2. Determine the critical $$ t $$-values**  
Since the sample size is $$ n = 67 $$, the degrees of freedom are  
$$
df = n - 1 = 66
$$

- For a $$ 90\% $$ confidence interval, the two-tailed significance level is $$ \alpha = 0.10 $$ so that each tail has $$ \frac{\alpha}{2} = 0.05 $$. The corresponding $$ t $$-value is approximately  
  $$
  t_{0.05,66} \approx 1.67
  $$

- For a $$ 95\% $$ confidence interval, $$ \alpha = 0.05 $$ and each tail has $$ \frac{\alpha}{2} = 0.025 $$. The corresponding $$ t $$-value is approximately  
  $$
  t_{0.025,66} \approx 2.00
  $$

**Step 3. Compute the margins of error**

- For the $$ 90\% $$ confidence interval:  
  $$
  \text{Margin of Error}_{90\%} = t_{0.05,66} \times SE \approx 1.67 \times 0.635 \approx 1.06
  $$

- For the $$ 95\% $$ confidence interval:  
  $$
  \text{Margin of Error}_{95\%} = t_{0.025,66} \times SE \approx 2.00 \times 0.635 \approx 1.27
  $$

**Step 4. Construct the confidence intervals**

- The $$ 90\% $$ confidence interval is given by  
  $$
  \bar{x} \pm \text{Margin of Error}_{90\%} = 17.5 \pm 1.06
  $$  
  Thus, the interval is approximately  
  $$
  (17.5 - 1.06,\; 17.5 + 1.06) = (16.44,\; 18.56)
  $$

- The $$ 95\% $$ confidence interval is given by  
  $$
  \bar{x} \pm \text{Margin of Error}_{95\%} = 17.5 \pm 1.27
  $$  
  Thus, the interval is approximately  
  $$
  (17.5 - 1.27,\; 17.5 + 1.27) = (16.23,\; 18.77)
  $$

**Answer:**  
- $$90\%$$ confidence interval: $$ (16.44,\; 18.56) $$ contacts/week  
- $$95\%$$ confidence interval: $$ (16.23,\; 18.77) $$ contacts/week
The $$90\%$$ confidence interval for the population mean number of weekly customer contacts is $$16.44$$ to $$18.56$$ contacts/week. The $$95\%$$ confidence interval is $$16.23$$ to $$18.77$$ contacts/week.

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