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DedPeter [7]
4 years ago
11

A brochure claims that the average maximum height for a certain type of plant is 0.7 m. A gardener suspects that this is not acc

urate locally due to variation in soil conditions, and believes the local height is shorter. A random sample of 40 mature plants is taken. The mean height of the sample is 0.65 m with a standard deviation of 0.20 m. Test the claim that the local mean height is less than 0.7 m using a 5% level of significance.
Mathematics
1 answer:
larisa [96]4 years ago
4 0

Answer:

As Z, it is possible to reject null hypotesis. It means that the local mean height is less tha 0.7 m with a 5% level of significance.

Step-by-step explanation:

1. Relevant data:

\mu=0.70\\N=40\\\alpha=0.05\\X=0.65\\s=0.20

2. Hypotesis testing

H_{0}=\mu=0.70

H_{1} =\mu< 0.70

3. Find the rejection area

From the one tail standard normal chart, whe have Z-value for \alpha=0.05 is 1.56

Then rejection area is left 1.56 in normal curve.

4. Find the test statistic:

Z=\frac{X-\mu_{0} }{\sigma/\sqrt{n}}

Z=\frac{0.65-0.70}{0.20/\sqrt{40}}\\Z=-1.58

5. Hypotesis Testing

Z_{\alpha}=1.56\\Z=-1.58

-1.58

As Z, it is possible to reject null hypotesis. It means that the local mean height is less tha 0.7 m with a 5% level of significance.

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Step-by-step explanation: 105 divided by 15 = 7

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3 years ago
You run around the perimeter of a baseball field at a rate of at least 10 feet per second. Which of the following are possible a
vovikov84 [41]
1) we have to calculate the lenght of the baseball field.

perimeter of the baseball field=2(radius)+lenght of a semicircle.

2(radius)=2(175 ft)=350 ft
lenght of a semicircle=40% of a circumference=(40/100)(2πr)

Perimeter of the baseball field=2(175 ft) + (40/100)(2π175 ft)=
=350 ft + 140π ft=350 ft + 439.82 ft=789.82 ft,


2)Now, we calculate the amounts of time that it takes you to ruan arount the baseball field.

10 ft------------------------1 second
789,82 ft-----------------    x

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answer: 79 seconds.
3 0
3 years ago
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Debbie is running to raise money for a fundraiser. If her goal is to run 56 miles , how many days will debbie need to run if she
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Answer:

32 days.

Step-by-step explanation:

56 / 1.75 = 56 / (7/4) = 56 * 4 / 7 = 32.

3 0
4 years ago
What is 5+2 x 1-3+8 and 2+5 x 1
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Answer:

Step-by-step explanation:

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3 years ago
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Suppose a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.00
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Answer:

We conclude that the standard deviation of the diameter of a golf ball is less than 0.005 inch.

Step-by-step explanation:

We are given that a professional golfing association requires that the standard deviation of the diameter of a golf ball be less than 0.005 inch.

Assume that the population is normally distributed: 1.678, 1.681, 1.676, 1.684, 1.676, 1.679, 1.681, 1.681, 1.677, 1.676, 1.681, 1.683.

Let \sigma = <u><em>population standard deviation of the diameter of a golf ball.</em></u>

SO, Null Hypothesis, H_0 : \sigma \geq 0.005 inch     {means that the standard deviation of the diameter of a golf ball is more than or equal to 0.005 inch}

Alternate Hypothesis, H_0 : \sigma < 0.005 inch     {means that the standard deviation of the diameter of a golf ball is less than 0.005 inch}

The test statistics that would be used here <u>One-sample Chi-square test statistics</u>;

                            T.S. =  \frac{(n-1)s^{2} }{{\sigma^{2} }}  ~ \chi^{2} __n_-_1

where, s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = 0.00281 inch

           n = sample size = 12

So, <u><em>the test statistics</em></u>  =  \frac{(12-1)\times 0.00281^{2} }{{0.005^{2} }}  ~ \chi^{2} __1_1

                                     =  3.47

The value of chi-square test statistics is 3.47.

Also, the P-value of test statistics is given by the following formula;

                P-value = P( \chi^{2} __1_1 < 3.47) = 0.0182

Since, the P-value of the test statistics is less than the level of significance as 0.0182 < 0.05, so we reject our null hypothesis.

<u>Now, at 0.05 significance level the chi-square table gives critical value of 4.575 at 11 degree of freedom for left-tailed test.</u>

Since our test statistic is less than the critical value of chi-square as 3.47 < 4.575, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that the standard deviation of the diameter of a golf ball is less than 0.005 inch.

7 0
3 years ago
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