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Alex17521 [72]
3 years ago
8

At a college the scores on the chemistry final exam are approximately normally distributed, with a mean of 77 and a standard dev

iation of 10. The scores on the calculus final are also approximately normally distributed, with a mean of 83 and a standard deviation of 14. A student scored 81 on the chemistry final and 81 on the calculus final. Relative to the students in each respective class, in which subject did the student do better?
Mathematics
1 answer:
lbvjy [14]3 years ago
7 0

Answer:

Student do better in Chemistry subject.

Step-by-step explanation:

We are given that at a college the scores on the chemistry final exam are approximately normally distributed, with a mean of 77 and a standard deviation of 10. The scores on the calculus final are also approximately normally distributed, with a mean of 83 and a standard deviation of 14.

A student scored 81 on the chemistry final and 81 on the calculus final.

And we have to find that  in which subject did the student do better.

<em>Firstly, Let X = scores on the chemistry final exam</em>

<em>So, X ~ N(</em>\mu=77,\sigma^{2} = 10^{2}<em>)</em>

<em>Also, let Y = scores on the calculus final exam</em>

<em>So, Y ~ N(</em>\mu=83,\sigma^{2} = 14^{2}<em>)</em>

For finding in which subject did the student do better, we will find the z score for both the exams of student because the higher the z score, the better is the student perform in that exam.

The z-score probability distribution is given by;

              Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = mean score for respective subjects

            \sigma = standard deviation

  • <u>The z-score of Chemistry final exam is calculated as;</u>

Since we are given the student score of 81 on the chemistry final exam,

So,  z-score = \frac{81-77}{10} = 0.4  {where \mu=77 and \sigma =10 }

  • <u>The z-score of Calculus final exam is calculated as;</u>

Since we are given the student score of 81 on the calculus final exam,

So,  z-score = \frac{81-83}{14} = -0.143     {where \mu=83 and \sigma =14 }

AS we can clearly see that the z score of Chemistry final exam is higher than that of Calculus exam so the student do better in Chemistry subject exam.

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Step-by-step explanation:

Given the terms ; 4, 12a, 6, 2a, and 24

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from the last term = 24 = 6 x 4 ...........(2)

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= 12(2) + 24 = 24 + 24 = 48 (LHS)

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Two major automobile manufacturers have produced compact cars with engines of the same size. We are interested in determining wh
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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 <u>Two-sample t test statistics</u> 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) <u><em>The test statistics</em></u>  =  \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 <u>we fail to reject our null hypothesis</u>.

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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