"Chips Ahoy! $1,000$ Chips Challenge." As reported by B. Warner and J. Rutledge in the paper "Checking the Chips Ahoy! Guarantee" (Chanee, Vol. 12. Issue 1. pp. 10-14), a random sample of forty-two $18-$ounce bags of Chips Ahoy! cookies yielded a mean of $1261.6$ chips per bag with a standard deviation of $117.6$ chips per bug.

a. Determine a $95\%$ confidence interval for the mean number of chips per bag for all $18-$ounce bags of Chips Ahoy! cookies, and interpret your result in words.

b. Can you conclude that the average $18-$ounce bag of Chips Ahoy! cookies contain at least $1000$ chocolate chips? Explain your answer.

A representative sample of $42$$18-$ounce bags of Chips The average cost of Ahoy! cookies were $1261.6$per bag, with a standard deviation of $117.6$per bug.

The formula used:$z=\overline{x}\pm t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)$

Calculate the $95\%$confidence interval for every Chips Ahoy! $18-$ounce plastic bags of cookies, and write down your interpretation.

Consider $\overline{x}=1,261.6,n=42,s=117.6$, and confidence level is $95\%$

From "Table IV Values of $t_{\alpha }$" the required value of $t_{\frac{\alpha }{2}}$for $95\%$confidence with $41(=42-1)$degrees of freedom is $2.020$

Thus, the confidence interval is,

$$\begin{gathered} \overline{x}\pm t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)=1,261.6\pm 2.020\left(\frac{117.6}{\sqrt{42}}\right) \\ =1,261.6\pm 2.020(18.146) \\ =1,261.6\pm 36.655 \\ =(1,224.945,1,298.255) \end{gathered}$$

In order to understand your answer in words, For all $18-$ounce packs of Chips, the $95$ percent confidence interval for the mean amount of chips per bag Ahoy! Cookies are delicious. $(1,224.945,1,298.255)$

Because the value $1,000$ does not appear in the interval obtained in section (a), it may be deduced that the average $18-$ounce package of Chips Ahoy! cookies contain at least$1,000$ chocolate chips (a).