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3 years ago

Alejandro bought a new refrigerator for $856 with a 25% discount and

Mathematics

2 answers:

Maurinko [17]3 years ago

8 0

Answer:

695.93

Step-by-step explanation:

75% of 856 is 642

642 + 53.93 = 695.93

6.3 = 53.93

oksano4ka [1.4K]3 years ago

7 0

Answer:

$695.94

Step-by-step explanation:

25% of 856 = 214

6.3% of 856 = 53.928

856 + 53.928 = 909.928

909.928 - 214 = 695.928 about 695.94

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Two major automobile manufacturers have produced compact cars with engines of the same size. We are interested in determining wh

Molodets [167]

Answer:

(A) The mean for the differences is 2.0.

(B) The test statistic is 1.617.

(C) At 90% confidence the null hypothesis should not be rejected.

Step-by-step explanation:

We are given that a random sample of eight cars from each manufacturer is selected, and eight drivers are selected to drive each automobile for a specified distance.

The following data (in miles per gallon) show the results of the test;

Driver Manufacturer A Manufacturer B

1 32 28

2 27 22

3 26 27

4 26 24

5 25 24

6 29 25

7 31 28

8 25 27

Let $\mu_1$ = mean MPG for the fuel efficiency of Manufacturer A brand

$\mu_2$ = mean MPG for the fuel efficiency of Manufacturer B brand

SO, Null Hypothesis, $H_0$ : $\mu_1-\mu_2=0$ or $\mu_1= \mu_2$ {means that there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq 0$ or $\mu_1\neq \mu_2$ {means that there is a significant difference in the mean MPG (miles per gallon) for the fuel efficiency of these two brands of automobiles}

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

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n__1+_n__2-2$

where, $\bar X_1$ = sample mean MPG for manufacturer A = $\frac{\sum X_A}{n_A}$ = 27.625

$\bar X_2$ = sample mean MPG for manufacturer B = $\frac{\sum X_B}{n_B}$ = 25.625

$s_1$ = sample standard deviation for manufacturer A = $\sqrt{\frac{\sum (X_A-\bar X_A)^{2} }{n_A-1} }$ = 2.72

$s_2$ = sample standard deviation manufacturer B = $\sqrt{\frac{\sum (X_B-\bar X_B)^{2} }{n_B-1} }$ = 2.20

$n_1$ = sample of cars selected from manufacturer A = 8

$n_2$ = sample of cars selected from manufacturer B = 8

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(8-1)\times 2.72^{2}+(8-1)\times 2.20^{2} }{8+8-2} }$ = 2.474

(A) The mean for the differences is = 27.625 - 25.625 = 2

(B) The test statistics = $\frac{(27.625-25.625)-(0)}{2.474 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ ~ $t_1_4$

= 1.617

(C) Now at 10% significance level, the t table gives critical values between -1.761 and 1.761 at 14 degree of freedom for two-tailed test. Since 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 due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is a not any significant difference in the mean MPG (miles per gallon) when testing for the fuel efficiency of these two brands of automobiles.

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