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Harrizon [31]
2 years ago
10

Stacy recorded the number of minutes she used her

Mathematics
1 answer:
Ostrovityanka [42]2 years ago
6 0

Using the t-distribution, the most appropriate conclusion for the hypotesis test is given by Phone use did not change.

<h3>How to find the value of z-statistic for population mean? (z test statistic)</h3>

Suppose we're specified that:

The sample mean = \overline{x}

The population mean =\mu

The population standard deviation = \sigma

The sample size = n

Then the z-statistic for this data is found as:

Z = \dfrac{\overline{x} - \mu}{\sigma/\sqrt{n}}

In this problem, the values of those parameters are as follows:

The sample mean  \overline{x} =  -0.2

The population mean =\mu = 0

The population standard deviation = \sigma = 9.1

The sample size = n = 200

Hence, the test statistic is given by:

Z = \dfrac{\overline{x} - \mu}{\sigma/\sqrt{n}}\\\\\\Z = \dfrac{-0.2- 0}{9.1/\sqrt{200}}\\\\z = 0.31

Thus,

t = 0.31.

Since the absolute value of the test statistic is less than the critical value, we do not reject the null hypothesis and the conclusion is:

Phone use did not change.

Learn more about z-statistic here:

brainly.com/question/1640298

#SPJ1

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andreyandreev [35.5K]

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A chemist wishes to mix a 30% acid solution with a 65% acid solution to obtain 175 quarts of 55% acid solution . How many quarts
Eva8 [605]

x = quarts of 30% acid solution

y = quarts of 65% solution

x + y = 175

Each quart of the 30% solution contributes 0.30x quarts of acid; similarly, each quart of the 65% solution contributes 0.65y quarts of acid. So

0.3x + 0.65y = 0.55*175 = 96.25

That is, the chemist wants 175 quarts of a new mixture with a 55% concentration of acid, which comes out to 95.25 total quarts of acid. The total amounts of acid from either sides of this equation must balance out.

Now solve for x and y:

x + y = 175  ==>  y = 175 - x

0.3x + 0.65y = 96.25  ==>  0.3x + 0.65(175 - x) = 96.25

==>  0.3x + 113.75 - 0.65x = 96.25

==>  17.5 = 0.35x

==>  x = 50  ==>  y = 125

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Needdd helppp pleasseeee
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