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An entertainment company owns and operates movie theaters in Wyoming. The president of the company is concerned that film rentals are hurting the business. They directed a staff member to estimate the total number of films rented by households in Wyoming in a particular month. A phone survey involving a random sample of 300 homes was conducted with the following results - xbar = 2.4 films. Sx = 1.6 films. Then the 90% confidence interval estimate for the total number of films rented by the 211,000 households in that particular month?

Here you want to get the area in the center of the normal distribution of 90%, leaving you }5% at the two tails. Using the table for the normal distribution, find the Z-value for .05.

Using Z = (X - mean)/SE (Not SD. See other post.), find the value of X. That value, subtracted and added to the sample mean, gives you the 90% confidence interval.

I hope this helps. Thanks for asking.

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