Step 1
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1. a. The confidence level is 98%. Therefore, the level of significance is 1 – confidence level = 2% or 0.02.
The sample size, n is provided as 144 whose average is 44.7. The standard deviation is 16.5.
The data is continuous and the sample is 144 which is greater than 30. Therefore, it is assumed that the data follows the normal distribution making the z distribution appropriate to obtain the confidence interval.
Step 2
b. The 98% confidence interval for population mean number of texts per day is,
\(\displaystyle{C}.{I}.=\overline{{{x}}}\pm{z}_{{\frac{{\alpha}}{{{2}}}}}{\left\lbrace{\frac{{{s}}}{{\sqrt{{{n}}}}}}\right\rbrace}\)

\(\displaystyle={44.7}\pm{z}_{{\frac{{{0.02}}}{{{2}}}}}{\frac{{{16.5}}}{{\sqrt{{{144}}}}}}\)

\(\displaystyle={44.7}\pm{z}_{{{0.01}}}{\frac{{{16.5}}}{{\sqrt{{{144}}}}}}\)

\(\displaystyle{44.7}\pm{\left({2.33}\right)}{\frac{{{16.5}}}{{\sqrt{{{144}}}}}}\) (Using standard normal table) \(\displaystyle={\left({41.496},{47.904}\right)}\) c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About 98% of these confidence intervals will contain the true population number of texts per day and about 2% (100 – 98) will not contain the true population mean number of texts per day.

\(\displaystyle={44.7}\pm{z}_{{\frac{{{0.02}}}{{{2}}}}}{\frac{{{16.5}}}{{\sqrt{{{144}}}}}}\)

\(\displaystyle={44.7}\pm{z}_{{{0.01}}}{\frac{{{16.5}}}{{\sqrt{{{144}}}}}}\)

\(\displaystyle{44.7}\pm{\left({2.33}\right)}{\frac{{{16.5}}}{{\sqrt{{{144}}}}}}\) (Using standard normal table) \(\displaystyle={\left({41.496},{47.904}\right)}\) c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About 98% of these confidence intervals will contain the true population number of texts per day and about 2% (100 – 98) will not contain the true population mean number of texts per day.