Question

an app estimates phone usage by counting the number of times a phone screen is unlocked during the course of a day. a sample of 25 users is shown. complete parts a thro
123 65 119 145 114
125 114 91 113 125
88 141 105 116 121
186 136 65 128 107
97 126 101 90 10

a. what is the mean and standard deviation?
the mean is 110.04
(round to the nearest hundredth as needed.)
the standard deviation is 32.72
(round to the nearest hundredth as needed.)
b. assuming the sample standard deviation matches the population standard deviation, what is the margin of error?
the margin of error is
(round to the nearest hundredth as needed.)

Explanation:

Step1: Recall Margin of Error Formula

For a sample with known population standard deviation (here, sample std dev is used as population std dev), and sample size \( n \), the margin of error \( E \) for a 95% confidence interval (common, assuming this unless stated otherwise) is \( E = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \). For 95% confidence, \( z_{\alpha/2} = 1.96 \).

Step2: Identify Values

We have \( \sigma = 32.72 \), \( n = 25 \).

Step3: Calculate Standard Error

Standard error \( = \frac{\sigma}{\sqrt{n}} = \frac{32.72}{\sqrt{25}} = \frac{32.72}{5} = 6.544 \).

Step4: Calculate Margin of Error

\( E = 1.96 \times 6.544 \approx 12.83 \).

Answer:

The margin of error is \( 12.83 \) (assuming 95% confidence level; if a different confidence level is intended, the \( z \)-value changes, but 95% is standard for such problems without specification).