Question

A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is normally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams. What is the​ 90% confidence interval for the true mean weight of​ orders?

Answer 1

Answer:

90% confidence interval for the true mean weight of​ orders is [103.86 , 116.14].

Step-by-step explanation:

We are given that Past records indicate that the weight of orders is normally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams.

Firstly, the pivotal quantity for 90% confidence interval for the true mean weight of​ orders is given by;

         P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 110 grams

             s = sample standard deviation = 14 grams

             n = sample of orders = 16

             $\mu$ = true population mean

Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.

So, 90% confidence interval for the true mean, $\mu$ is ;

P(-1.753 < $t_1_5$ < 1.753) = 0.90  {As the critical value of t at 15 degree of

                                                 freedom are -1.753 & 1.753 with P = 5%}

P(-1.753 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 1.753) = 0.90

P( $-1.753 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $1.753 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X -1.753 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +1.753 \times {\frac{s}{\sqrt{n} }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X -1.753 \times {\frac{s}{\sqrt{n} }$ , $\bar X +1.753 \times {\frac{s}{\sqrt{n} }$ ]

                                                 = [ $110 -1.753 \times {\frac{14}{\sqrt{16} }$ , $110 +1.753 \times {\frac{14}{\sqrt{16} }$ ]

                                                 = [103.86 , 116.14]

Therefore, 90% confidence interval for the true mean weight of​ orders is [103.86 , 116.14].

Answer 2

Answer:

90% confidence interval for the true mean weight of orders is between a lower limit of 103.8645 grams and an upper limit of 116.1355 grams.

Step-by-step explanation:

Confidence interval for true mean weight is given as sample mean +/- margin of error (E)

sample mean = 110 g

sample sd = 14 g

n = 16

degree of freedom = n - 1 = 16 - 1 = 15

confidence level = 90% = 0.9

significance level = 1 - C = 1 - 0.9 = 0.1 = 10%

critical value (t) corresponding to 15 degrees of freedom and 10% significance level is 1.753

E = t × sample sd/√n = 1.753×14/√16 = 6.1355 g

Lower limit of sample mean = sample mean - E = 110 - 6.1355 = 103.8645 g

Upper limit of sample mean = sample mean + E = 110 + 6.1355 = 116.1355 g

90% confidence interval is (103.8645, 116.1355)