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timofeeve [1]
3 years ago
11

A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random

samples of 5 babies' weights (pounds) for each gender showed the following:
Boys: 8.0 4.7 7.3 6.2 3.4
Girls: 5.3 2.8 6.4 6.8 7.4
To test the researcher's hypothesis, we should use the:

a) paired (dependent) samples t-test
b) independent samples t-test
c) large-sample z-test
d) t-test for correlation
Mathematics
1 answer:
Yuliya22 [10]3 years ago
7 0

Answer:

b) independent samples t-test

t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154  

p_v =2*P(t_{8}>0.154) =0.881  

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).  

Step-by-step explanation:

The system of hypothesis on this case are:  

Null hypothesis: \mu_1 = \mu_2  

Alternative hypothesis: \mu_1 \neq \mu_2  

Or equivalently:  

Null hypothesis: \mu_1 - \mu_2 = 0  

Alternative hypothesis: \mu_1 -\mu_2\neq 0  

Our notation on this case :  

n_1 =5 represent the sample size for group 1  

n_2 =5 represent the sample size for group 2  

We can calculate the mean and deviation for eaach group with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X_1 =5.92 represent the sample mean for the group 1  

\bar X_2 =5.74 represent the sample mean for the group 2  

s_1=1.88 represent the sample standard deviation for group 1  

s_2=1.81 represent the sample standard deviation for group 2  

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:  

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}  

And now we can calculate the statistic:  

t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154  

The degrees of freedom are given by:  

df=5+5-2=8

And now we can calculate the p value using the altenative hypothesis:  

p_v =2*P(t_{8}>0.154) =0.881  

So with the p value obtained and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).  

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A line contains the two points plotted in the coordinate plane below.
Andrew [12]

Answer:

The slope would be 3.

Step-by-step explanation:

The formula for finding slope is \frac{y1-y2}{x1-x2}
We can plug these numbers into the equation by doing \frac{1--5}{2-0}
1--5 is equal to 1+5, which is 6.
2-0 is equal to 2.
\frac{6}{2} is 3, which is how you get your answer.
I hope this helps! :)

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2 years ago
What is the equation of the line shown in this graph? <br><br>any links will be reported<br><br>​
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Answer:

y = 1

Step-by-step explanation:

Slope is written in the form y=mx+b. Where m is the slope and b is the y-intercept.

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Likewise, if the y-intercept is at (0,0), then you'd only write y equals the slope (y = mx).

Hopefully this makes sense :)

Btw, I also answered in the main comments earlier, so you can check that out as well.

7 0
2 years ago
Ben plans some turf to cover his triangular lawn which is 4.8 yards high. Its area measures 40.2 square yards. How much does the
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3 0
3 years ago
Leicester City Fanstore (LCF) will be selling the "new season jersey" for the 2018-2019 season. The regular price of the jersey
andrew-mc [135]

Complete Question

Leicester City Fanstore (LCF) will be selling the "new season jersey" for the 2018-2019 season. The regular price of the jersey is $80. Each jersey costs $40. Leftover jerseys will be sold at the end of the season (or later) at $30. Since jerseys are produced in China and lead time is long, Puma wants LCF to decide the quantity right now (December 2017).

a. After some analysis using historic data, LCF expects that the demand will follow a Normal distribution with a mean 40,000 and a standard deviation of 8,000 due to uncertainty in team performance. How many jerseys should LCF order?

b. If a customer cannot buy the jersey from LCF (in the case of a stock-out), they may leave the store disappointed and use other channels (such as Puma stores or puma.com) in future. LCF thinks that the lost customer goodwill is around $10. Should LCF change their decision in part (a)? If yes, please state the number of jerseys LCF should order.

c. Please state whether the following statement is always true, and give a brief explanation. If C_o =C_i, the news vendor solution is the mean.

Answer:

a

   N  =  46728

b

  n  =  47728

c

  Yes it is always true  

Step-by-step explanation:

From the question we are told that

   The regular price of the jersey is  P_r = \$ 80

    The cost of producing a jersey is C=  \$ 40

  The left-over price of the jersey is P_o  = $ 30

   The mean is  \mu =  40000

   The standard deviation is \sigma =  8000

   The cost of lost customer goodwill is C_g = \$ 10

Generally the fund that LCF will loss for one jersey if they order for too many  jersey (i.e more than they need )is mathematically represented  

           C_o  =  P_o -  C

=>        C_o  = 40 - 30

=>        C_o  = \$ 10

Generally the fund that LCF will loss for one jersey if they order lesser amount  jersey (i.e less than they need )is mathematically represented  

           C_i  =  P_r -  C

=>        C_i  = 80 - 40

=>        C_i  = \$ 40

Generally the critical ratio is mathematically represented as

             Z  =  \frac{C_i }{ C_i + C_o}

=>           Z =  \frac{40}{ 40 + 10}

=>           Z  = 0.8

Generally the critical value of  Z  = 0.8 to the right of the normal curve is

         z = 0.841

Generally the optimal quantity of jersey to order is mathematically represented as

             N  =  \mu  * [z *  \sigma]

=>          N  =  40000 * [0.841 *   8000]

=>          N  =  46728

Considering question b

  Generally considering the factor of customer goodwill  the fund that LCF will loss for one jersey if they order lesser amount  jersey (i.e less than they need )is mathematically represented  as

           C_k  =  C_i +  C_g

=>        C_k  =  40 +  10

=>        C_k  = \$ 50

Now   the critical ratio is mathematically represented as

             Z  =  \frac{C_k }{ C_k + C_o}

=>           Z =  \frac{50}{ 50 + 10}

=>           Z  = 0.833

Generally the critical value of  Z  = 0.833 to the right of the normal curve is

         z = 0.966

Generally the optimal quantity of jersey to order is mathematically represented as

             n  =  \mu  * [z *  \sigma]

=>          n  =  40000 * [0.966 *   8000]

=>          n  =  47728

Considering question c

    When C_o =C_i then

The critical ratio is mathematically represented as

             Z  =  \frac{C_k }{ C_k + C_k}

=>           Z =  \frac{1}{ 2}

=>           Z  = 0.5          

Generally the critical value of  Z  = 0.5 to the right of the normal curve is

         z = 0

So

The optimal quantity of jersey to order is mathematically represented as

             n  =  \mu  * [z *  \sigma]

=>          n  =  40000 * [0*   8000]

=>          n  =  40000 = \mu

Hence the statement in c is true        

4 0
2 years ago
RSM HOMEWORK HELP!!! PLEASE HELP ME FAST I'M REALLY STUCK!!
stepan [7]

Answer:

15

Step-by-step explanation:

Consider right triangle EFD. The sum of the measures of all interior angles in triangle EFD is 180°, so

m\angle FED=180^{\circ}-90^{\circ}-3x=90^{\circ}-3x

Consider right triangle EBC. The sum of the measures of all interior angles in triangle EBC is 180°, so

m\angle BCE=180^{\circ}-(90^{\circ}-3x)-90^{\circ}=3x

Consider triangle ACD. The sum of the measures of all interior angles in triangle ACD is 180°, so

x+8x+3x=180^{\circ}\\ \\12x=180^{\circ}\\ \\x=15

3 0
3 years ago
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