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svetoff [14.1K]
2 years ago
11

Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

igma(1) = 0.010 liter and sigma(2) = 0.015 liter, respectively. A random sample of n(1) = 25 bottles from machine 1 and n(2) = 20 bottles from machine 2 results in average net contents of x-bar = 2.04 liters and x-bar = 2.07 liters.
(a) Test the hypothesis that both machines fill to the same net contents, using alpha = 0.05. What are your conclusions?
(b) Find the P-value for this test.
(c) Construct a 95% confiedence interval on the difference in mean fill volume.
Mathematics
1 answer:
stellarik [79]2 years ago
8 0

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.  

Then we have n_{1} = 25, \bar{x}_{1} = 2.04, \sigma_{1} = 0.010 and n_{2} = 20, \bar{x}_{2} = 2.07, \sigma_{2} = 0.015. The pooled estimate is given by  

\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative).  

The test statistic is T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}} and the observed value is t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257. T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value t_{0} falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by 2P(T0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}, i.e.,

-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}

where t_{0.025} is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

-0.03\pm(2.0167)(0.012459)(0.3), i.e.,

(-0.0225, -0.0375)

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Step-by-step explanation:

Converting percent to decimal : 2.2/100 = 0.022

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Finding how much it was worth after 11 years :

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2 years ago
<img src="https://tex.z-dn.net/?f=%2826%20%5Cdiv%20100%29%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5Ctimes%2010" id
taurus [48]

Answer:

\boxed{\bf \:  \cfrac{13}{5}}

<u>Or in Decimal:</u>

\boxed{\bf \: 2.6}

Step-by-step explanation:

<u>Given expression :-</u>

\sf \: ( 26 \div 100) \times 10

<u>Solution :-</u>

\sf  = (26 \div 100 )\times 10

This arithmetic expression may be rewritten as ;

\sf  =  \cfrac{26}{100}  \times 10

Step 1 : <u>Cancel the zero of 10 and one zero of 100</u> :-

\sf  =  \cfrac{26}{10 \cancel0}  \times 1 \cancel0

<em>Results to;</em>

\sf  =  \:  \cfrac{26}{10}  \times 1

\sf  =  \:  \cfrac{26}{10}

Step 2: <u>Cancel 26 and 10</u><u> </u><u>by 2</u> :-

\sf  =  \cfrac{ \cancel{26}}{ \cancel{10}}

<em>Results to;</em>

\sf = \cfrac{ \cancel{26} {}^{13} }{ \cancel{10} {}^{5} }

\sf  =  \cfrac{13}{5}

<em>It can also be in Decimal.</em>

That is;

\sf = 2.6

Hence, the answer of the expression would be 13/5 or 2.6 .

\rule{225pt}{2pt}

I hope this helps!

Let me know if you have any questions.

I am joyous to help!

3 0
2 years ago
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(n+9) / (-4) = - 1, or   (n+9) / 4 = 1, or n+9 = 4.  Then n = -5.

7 0
2 years ago
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