The weights of four randomly and independently selected bags of tomatoes labeled 5.0 pounds were found to be 5.1, 5.0, 5.4, and 5.5 pounds. Assume Normality. Answer parts (a) and (b) below. a. Find a 95% confidence interval for the mean weight of all bags of tomatoes. (nothing comma nothing )(Type integers or decimals rounded to the nearest hundredth as needed. Use ascending order.) b. Does the interval capture 5.0 pounds? Is there enough evidence to reject a mean weight of 5.0 pounds? A. The interval does not capture 5.0 pounds, so there is enough evidence to reject a mean weight of 5.0 pounds. It is not plausible the population mean weight is 5.0 pounds. B. The interval captures 5.0 pounds, so there is enough evidence to reject a mean weight of 5.0 pounds. It is not plausible the population mean weight is 5.0 pounds. C. The interval does not capture 5.0 pounds, so there not is enough evidence to reject a mean weight of 5.0 pounds. It is plausible the population mean weight is 5.0 pounds. D. The interval captures 5.0 pounds, so there is not enough evidence to reject a mean weight of 5.0 pounds. It is plausible the population mean weight is 5.0 pounds.

(b) The interval captures 5.0 pounds, so there is not enough evidence to reject a mean weight of 5.0 pounds. It is plausible the population mean weight is 5.0 pounds.

Step-by-step explanation:

We are given that the weights of four randomly and independently selected bags of tomatoes labeled 5.0 pounds were found to be 5.1, 5.0, 5.4, and 5.5 pounds.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean weight of bags of tomatoes = [tex]\frac{\sum X}{n}[/tex] = 5.25 pounds

s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex] = 0.24 pounds

n = sample of bags = 4

[tex]\mu[/tex] = population mean weight of all bags of tomatoes

Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 95% confidence interval for the population mean weight, [tex]\mu[/tex] is ;

P(-3.182 < [tex]t_3[/tex] < 3.182) = 0.95 {As the critical value of t at 3 degree of

freedom are -3.182 & 3.182 with P = 2.5%}

P(-3.182 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 3.182) = 0.95

P( [tex]-3.182 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]3.182 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

P( [tex]\bar X-3.182 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+3.182 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95

95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-3.182 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+3.182 \times {\frac{s}{\sqrt{n} } }[/tex] ]

= [ [tex]5.25-3.182 \times {\frac{0.24}{\sqrt{4} } }[/tex] , [tex]5.25+3.182 \times {\frac{0.24}{\sqrt{4} } }[/tex] ]

= [4.87 pounds , 5.63 pounds]

(a) Therefore, 95% confidence interval for the mean weight of all bags of tomatoes is [4.87 pounds , 5.63 pounds].

(b) Yes, the above confidence capture 5.0 pounds as it will lie within the two value of 4.87 pounds and 5.63 pounds.

Since, the interval captures 5.0 pounds, so there is not enough evidence to reject a mean weight of 5.0 pounds. It is plausible the population mean weight is 5.0 pounds.