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klasskru [66]
3 years ago
5

Mr.Blankenship has 3/8 pound of candy to give to the top 3 finalists in the spelling bee if each finalist receives an equal amou

nt, how many pounds of candy will each finalist receive?
Mathematics
1 answer:
Triss [41]3 years ago
6 0

Answer:

3/8

Step-by-step explanation:

3/8+8/9

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An engineer in charge of water rationing for the U.S. Army wants to determine if the average male soldier spends less time in th
lina2011 [118]

Answer:

a) H0: μm = μf versus Ha: μm < μf

b) t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200

c) p_v =P(t_{133}  

d) Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.

Step-by-step explanation:

Data given and notation  

\bar X_{m}=2.68 represent the mean for the sample male

\bar X_{f}=2.7 represent the mean for the sample female

s_{m}=0.65 represent the sample standard deviation for the males

s_{f}=0.5 represent the sample standard deviation for the females  

n_{m}=66 sample size for the group male  

n_{f}=69 sample size for the group female  

t would represent the statistic (variable of interest)  

Part a

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the average male soldier spends less time in the shower than the average female soldier, the system of hypothesis would be:  

Null hypothesis:\mu_{m}-\mu_{f}\geq 0  

Alternative hypothesis:\mu_{m} - \mu_{f}< 0  

Or equivalently:

Null hypothesis:\mu_{m}-\mu_{f}= 0  

Alternative hypothesis:\mu_{m} - \mu_{f}< 0  

And the best option is:

H0: μm = μf versus Ha: μm < μf

Part b

We don't have the population standard deviation, so for this case is better apply a t test to compare means, and the statistic is given by:  

t=\frac{(\bar X_{m}-\bar X_{f})-\Delta}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{f}}{n_{f}}}} (1)

And the degrees of freedom are given by df=n_m +n_f -2=66+69-2=133  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

What is the test statistic?

With the info given we can replace in formula (1) like this:  

t=\frac{(2.68-2.7)-0}{\sqrt{\frac{0.65^2}{66}+\frac{0.5^2}{69}}}}=-0.200

Part c What is the p-value?

Since is a left tailed test the p value would be:  

p_v =P(t_{133}  

Part d

The significance level given is \alpha =0.1 since the p value is higher than the significance level we can conclude:

Fail to reject the claim that the average shower times are the same for male and female soldiers because the P-value is greater than 0.1.

4 0
3 years ago
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