Mathematics, 29.11.2019 01:31 vperez5765

Let x equal the weight in grams of a "52-gram" snack pack of candies. assume that the distribution of times is n(mu, 6). a random sample of n = 10 observations of x yielded the following data: 55.95 56.54 57.58 55.13 57.48 56.06 59.93 58.30 52.57 58.46
a. give a point estimate for mu.
b. find the endpoints for a 95% confidence interval for mu. lower
c. on the basis of these very limited data, what is the probability that an individual snack pack selected at random is filled with less than 52 grams of candy?

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Answer from: zoe41

a) $\bar x= 56.8$

b) The 95% confidence interval is given by (55.282;58.318)

c) P(X

Step-by-step explanation:

1) Notation and definitions

n=10 represent the sample size

$\bar X$ represent the sample mean

represent the sample standard deviation

$\sigma^2= 6$

m represent the margin of error

Confidence =95% or 0.95

A 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".

2) Calculate the mean (Point of estimate for $\mu$ ) and standard deviation for the sample

On this case we need to find the sample standard deviation with the following formula:

$s=\sqrt{\frac{\sum_{i=1}^15 (x_i -\bar x)^2}{n-1}}$

And in order to find the sample mean we just need to use this formula:

$\bar x =\frac{\sum_{i=1}^{15} x_i}{n}$

The sample mean obtained on this case is $\bar x= 56.8$ and the deviation s=2.052. Since we know the population standard deviation $\sigma^2=6$ and $\sigma=2.449$

3) Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ .

We can find the critical values in excel using the following formulas:

"=NORM.INV(0.025,0,1)" for $t_{\alpha/2}=-1.96$

"=NORM.INV(1-0.025,0,1)" for $t_{1-\alpha/2}=1.96$

The critical value $zc=\pm 1.96$

3) Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

$m=z_c \frac{\sigma}{\sqrt{n}}$

$m=1.96 \frac{2.449}{\sqrt{10}}=1.518$

4) Calculate the confidence interval

The interval for the mean is given by this formula:

$\bar X \pm z_{c} \frac{\sigma}{\sqrt{n}}$

And calculating the limits we got:

$56.8 - 1.96 \frac{2.449}{\sqrt{10}}=55.282$

$56.8 + 1.96 \frac{2.449}{\sqrt{10}}=58.318$

The 95% confidence interval is given by (55.282;58.318)

On the basis of these very limited data, what is the probability that an individual snack pack selected at random is filled with less than 52 grams of candy?

On this case we can use as point of estimate for the mean the result from part a, $\bar x= 56.8$ and the variance is $\sigma^2 =6$ , so $\sigma =\sqrt{6}$ . And we are interested on this probability: