Nielsen media research wants to estimate the mean amount of time, in minutes, that full-time college students spend texting each weekday. find the sample size necessary to estimate that mean with a 15 minute margin of error. assume that a 96% confidence level is desired and that the standard deviation is estimated to be 112.2 minutes.

n=237

Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

We know that the margin of error for a confidence interval is given by:

  (1)

The next step would be find the value of , and

Using the normal standard table, excel or a calculator we see that:

If we solve for n from formula (1) we got:

And we have everything to replace into the formula:

And if we round up the answer we see that the value of n to ensure the margin of error required is n=237.

i think there was a typo. you wrote pat brings 3 quarts of ice-cream to the party. each serving of ice-cream is 1 6 quart. how many total servings of ice-cream does pat bring to the party? i think the bold is supposed to be 1/6. so 1 quart of ice cream = 6 servings. 6 * 3 = 18.

idk

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