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AfilCa [17]
3 years ago
9

Body armor provides critical protection for law enforcement personnel, but it does affect balance and mobility. The article "Imp

act of Police Body Armour and Equipment on Mobility" (Applied Ergonomics, 2013: 957–961) reported that for a sample of 52 male enforcement officers who underwent an acceleration task that simulated exiting a vehicle while wearing armor, the sample mean was 1.95 sec, and the sample standard deviation was .20 sec.
Does it appear that true average task time is less than 2 sec? Carry out a test of appropriate hypotheses using a significance level of .01.
Mathematics
1 answer:
yKpoI14uk [10]3 years ago
3 0

Answer:

No, there is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that true average task time with armor is less than 2 seconds.

Then, the null and alternative hypothesis are:

H_0: \mu=2\\\\H_a:\mu< 2

The significance level is 0.01.

The sample has a size n=52.

The sample mean is M=1.95.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.2.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.2}{\sqrt{52}}=0.028

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1.95-2}{0.028}=\dfrac{-0.05}{0.028}=-1.803

The degrees of freedom for this sample size are:

df=n-1=52-1=51

This test is a left-tailed test, with 51 degrees of freedom and t=-1.803, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0.039) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

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Wesley has 480 stamps in his collection. He puts these stamps into display cases. Each display
disa [49]

Answer:

Wesley needs 32 displays for his stamp collection

Step-by-step explanation:

480 stamps in total.

15 stamps in each case

480/ 15= 32

6 0
3 years ago
5) One restaurant charges $7 per person and $1 admission fee. A second compant Charges $5 per
Assoli18 [71]

The people will there need to be in order for the second company to be less expensive in 1st restaurant.

Always assume the unknowns are variables when attempting to find them, and then construct the equation using the parameters specified in the question.

Admission fee for 1st restaurant = $1

Charge for 1st restaurant per person = $7

Admission fee for 2nd restaurant = $15

The number of employees required for 2nd restaurant need to have in order to be less expensive = x

Total amount needed for 1st restaurant = $1 + $7 = $8

To make the 2nd restaurant less expensive =

As the total amount of 1st restaurant is $8 and the admission fee 2nd restaurant itself is $15 2nd restaurant is more expensive than the 1st one.

To learn more about Word problems related to arithmetic operation from the link:

brainly.com/question/24464054

#SPJ9

5 0
1 year ago
A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the
a_sh-v [17]

Answer:

Probability that the sample proportion will be less than 0.03 is 0.10204.

Step-by-step explanation:

We are given that a courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the true proportion is 0.04.

Also, 469 are sampled.

<em>Let </em>\hat p<em> = sample proportion</em>

The z-score probability distribution for sample proportion is given by;

              Z = \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion

           p = true proportion = 0.04

           n = sample size = 469

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the sample proportion will be less than 0.03 is given by = P( \hat p < 0.03)

       P( \hat p < 0.03) = P( \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < \frac{0.03-0.04}{\sqrt{\frac{0.03(1-0.03)}{469} } } ) = P(Z < -1.27) = 1 - P(Z \leq 1.27)

                                                                      = 1 - 0.89796 = 0.10204

Now, in the z table the P(Z \leq x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.27 in the z table which has an area of 0.89796.

Therefore, probability that the sample proportion will be less than 0.03 is 0.10204.

5 0
3 years ago
A cereal box is 8 in by 2in by 12in what is the surface area.​
LUCKY_DIMON [66]

192 in2

The formula for surface area is 2(lw + wh + hl) = 2(2*8 + 12*8 + 12*2) = 2(16+56+24) = 2(96) = 192

4 0
3 years ago
If you sell 30 items at $200 a piece and your commission rate is 20%, how much commission do
jarptica [38.1K]

Answer:

$1200

Step-by-step explanation:

There are 2 ways I immediately think about to solving the problem.

1) You multiply 30 and 200 , then multiply the product by .2 (i.e. 20%)

This way, you'll find out what the total amount was($6,000), then you multiply it by your commission rate.(.2 or 20%) to find your total profit($1,200).

2) You multiply 200 by .20 , then multiply the product by 30.

This way, you'll find out what your commission would be for a single item($40), then multiply by the total items sold (30) to find your total profit($1,200).

5 0
3 years ago
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