Question

Asked Nov 15, 2019

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A smart phone manufacturer is interested in constructing a 90% confidence interval for the proportion of smart phones that break before the warranty expires. 85 of the 1594 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible.

a. With 90% confidence the proportion of all smart phones that break before the warranty expires is between and .

b. If many groups of 1594 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about percent will not contain the true population proportion.

Step 1

**Confidence interval:**

The (1-α) % confidence interval for proportion *p *is given by,

Step 2

**Computing the 90% confidence interval for population proportion:**

It was found that, 85 out of 1,594 smart phones broke before the warranty expires.

That is, *n*= 1,594 and the sample proportion, *p-*hat* = *0.05332 (=85/1,594). The confidence level is 90%. Therefore, the significance level, α = 0.1 and the *z* value corresponding to α/2 is 1.645. The 90% confidence interval for population proportion is given by:

Step 3

**A).**

With 90% confidence, the proportion of all smart phones that break before the warranty expires is bet...

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