Dave the jogger runs the same route every day (about 2.2 miles). On 18 consecutive days, he recorded the number of steps on his Fitbit. The results were 3,450 3,363 3,228 3,360 3,304 3,407 3,324 3,365 3,290 3,289 3,346 3,252 3,237 3,210 3,140 3,220 3,103 3,129 Click here for the Excel Data File (a) Construct a 95% confidence interval for the true mean number of steps Dave takes on his run. (Round your answers to 2 decimal places.) The 95% confidence interval is from to . (b) What sample size would be needed to obtain an error of ± 20 steps with 95% confidence? (Enter your answer as a whole number (no decimals). Use a z-value taken to 3 decimal places in your calculations.) Sample size

Answer:a) The 95% confidence interval would be given by (3229.95;3326.49)   b) n=91Step-by-step explanation:1) Previous conceptsA confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".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".[tex]\bar X[/tex] represent the sample mean for the sample [tex]\mu[/tex] population mean (variable of interest)s represent the sample standard deviationn represent the sample size 2) Part aThe confidence interval for the mean is given by the following formula:[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1) In order to calculate the mean and the sample deviation we can use the following formulas: [tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  [tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  The mean calculated for this case is [tex]\bar X=3278.222[/tex]The sample deviation calculated [tex]s=97.054[/tex]In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:[tex]df=n-1=18-1=17[/tex]Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,17)".And we see that [tex]t_{\alpha/2}=2.110[/tex]Now we have everything in order to replace into formula (1):[tex]3278.222-2.110\frac{97.054}{\sqrt{18}}=3229.95[/tex]   [tex]3278.222+2.110\frac{97.054}{\sqrt{18}}=3326.49[/tex]So on this case the 95% confidence interval would be given by (3229.95;3326.49)   3) Part bThe margin of error is given by this formula:[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]    (4)And on this case we have that ME =+20 and we are interested in order to find the value of n, if we solve n from equation (4) we got:[tex]n=(\frac{z_{\alpha/2} s}{ME})^2[/tex]   (5)The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got [tex]z_{\alpha/2}=1.960[/tex], replacing into formula (5) we got:[tex]n=(\frac{1.960(97.054)}{20})^2 =90.46 \approx 91[/tex]So the answer for this case would be n=91 rounded up to the nearest integer