Find the opposite integer of 6

A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su

Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let $\mu_1$ = mean sales per market in Region 1.

$\mu_2$ = mean sales per market in Region 2.

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, $H_A$ : > $\mu_1-\mu_2\neq$ 0 {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;

T.S. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean sales in Region 1 = 84

$\bar X_2$ = sample mean sales in Region 2 = 78.3

$s_1$ = sample standard deviation of sales in Region 1 = 6.6

$s_2$ = sample standard deviation of sales in Region 2 = 8.5

$n_1$ = sample of supermarkets from Region 1 = 12

$n_2$ = sample of supermarkets from Region 2 = 17

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times 8.5^{2} }{12+17-2} }$ = 7.782

So, the test statistics = $\frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }$ ~ $t_2_7$

= 1.943

The value of t-test statistics is 1.943.

Now, at a 0.02 level of significance, the t table gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0

2 years ago

One more math help needed question #5

crimeas [40]

75% becuase it is more than 5o% by the looks of it

4 0

2 years ago

The body temperatures of adults are normally distributed with a mean of 98.6degrees° F and a standard deviation of 0.60degrees°

Schach [20]

Answer:

97.72% probability that their mean body temperature is greater than 98.4degrees° F.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$