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Andre45 [30]
2 years ago
7

Daphne likes to ski at a resort that is open from December through April. According to a sign at the resort, 20, percent of the

snow falls occur in December, 25, percent in January, 20, percent in February, 20, percent in March, and 15, percent in April. She wondered if the snow falls in her hometown followed this distribution, so she took a random sample of 80 days between December and April with snowfall and recorded their months. Here are her results:_______.
Month December January February March April
Days 16 11 16 18 19
She wants to use these results to carry out a x2 goodness-of-fit test to determine if the distribution of snowfalls in her hometown disagrees with the claimed percentages.
Mathematics
1 answer:
Arturiano [62]2 years ago
6 0

Answer:

Step-by-step explanation:

From the given information:

We can compute the  null hypothesis & the alternative hypothesis as:

{H_o}:\text{Distribution of snowfalls in her hometown is similar to claimed percentage }

{H_a}:\text{Distribution of snowfalls in her hometown is not similar to claimed percentage }

The degree of freedom = n - 1

The degree of freedom = 5 - 1

The degree of freedom = 4

At the level of significance of 0.05 and degree of freedom 4,

the rejection region = 9.488

However, we can compute the chi-square X² goodness of fit test as:

   

months  frequency (p)  observed O Expected E  Chi-square X^2= \dfrac{(O-E)^2}{E}

Dec          0.2                  16                   16                \dfrac{(16-16)^2}{16} =0      

Jan           0.250             11                   20                \dfrac{(11-20)^2}{20} =4.050      

Feb           0.200             16                  16                 \dfrac{(16-16)^2}{16} =0      

Mar           0.200             18                  16                 \dfrac{(18-16)^2}{16} =0.250

Apr           0.150               19                  12                 \dfrac{(19-12)^2}{12} =4.083

Total            1.000           80                 80                                  8.3833    

∴

The test statistics X² = 8.3833

Thus; we fail to reject the H_o since test statistics X² doesn't fall in the rejection region.

Therefore; there is sufficient evidence to conclude that the distribution of snowfalls in her hometown is not similar to the claimed percentage.

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An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were
vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is  significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

<u>Step:-(1)</u>

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

<u>Step :-2</u>

<u>Null hypothesis: H₀:</u> there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

<u>Alternative hypothesis :H₁: </u>there is  significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

                     Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} }  } }

Given  n₁=n₂=60.

                    Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60}  } }

On calculation, we get

                   Z =   \frac{4}{\sqrt{0.6833} }

                   z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is  significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

3 0
2 years ago
Silva set up a dog-walking business. He earned $60 in June. He is projected to earn 110% more in July. How much is Silva project
snow_lady [41]

9514 1404 393

Answer:

  $126

Step-by-step explanation:

110% more than 100% is 210% of his June earnings:

  july = 2.10×$60 = $126

Silva is projected to make $126 in July.

_____

As always, you can think of % and /100 as being interchangeable.

  210% = 210/100 = 2.10

7 0
2 years ago
5 bricklayers can lay a total of 50 bricks in 30 minutes. How many bricklayers will
iragen [17]

Answer:

First look at the number of bricks alone.

Going from 50 bricks to 60 bricks is more work, thus it will require more people. The number of people would be the ratio of the 2. Since the number must be larger, you know the numerator must be the larger of the 2 numbers, so you get 60/50

Next look at the time alone.

Going from 30 minutes to 20 minutes is more work, thus it will require more people. The number of people would be the ratio of the 2. Since the number must be larger, you know the numerator must be the larger of the 2 numbers, so you get 30/20

Now you can just multiply everything.

= 5*60/50*30/20

= 5*6/5*3/2

= 90\10

= 9.

8 0
2 years ago
Find the first partial derivatives of the function. (sn = x1 + 2x2 + ... + nxn; i = 1, ..., n. give your answer only in terms of
liq [111]

Answer:

  ∂u/∂xi = i·cos(sn)

Step-by-step explanation:

For u = sin(v), the partial derivative of u with respect to xi is ...

  ∂u/∂xi = cos(v)·∂v/xi

In this case, v=sn, and ∂sn/∂xi = i, so the derivatives of interest are ...

  ∂u/∂xi = i·cos(sn)

6 0
2 years ago
What is the scientific notation of 0.0121
jek_recluse [69]
You move to the right, so the exponent is gonna be negative.
1.21 * 10-²
6 0
2 years ago
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