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tigry1 [53]
3 years ago
12

In an experiment to determine whether there is a systematic difference between the weights obtained with two different mass bala

nces, six specimens were weighed, in grams, on each balance. The following data were obtained:
Specimen A B
1 13.76 13.74
2 12.47 12.45
3 10.09 10.08
4 8.91 8.92
5 13.57 13.54
6 12.74 12.75
Can you conclude that the mean weight differs between the two balances?
i). State the null and alternative hypotheses.
ii). Compute the test statistic.
iii). State a conclusion using the a =0.05 level of significance.
Mathematics
1 answer:
Maslowich3 years ago
3 0

Answer:

H0: μd=0 Ha: μd≠0

t= 0.07607

On the basis of this we conclude that the mean weight differs between the two balances.

Step-by-step explanation:

The null and alternative hypotheses as

H0: μd=0 Ha: μd≠0

Significance level is set at ∝= 0.05

The critical region is t ( base alpha by 2 with df=5) ≥ ± 2.571

The test statistic under H0 is

t = d/ sd/ √n

Which has t distribution with n-1 degrees of freedom

Specimen        A               B           d = a - b         d²

1                     13.76        13.74         0.02           0.004

2                    12.47        12.45          0.02         0.004

3                    10.09        10.08           0.01        0.001

4                       8.91       8.92          -0.01          0.001

5                     13.57      13.54           0.03        0.009

<u>6                     12.74      12.75          -0.01        0.001</u>

<u>∑                                                      0.06         0.0173</u>

d`= ∑d/n= 0.006/6= 0.001

sd²= 1/6( 0.0173- 0.006²/6) = 1/6 ( 0.017294) = 0.002882

sd= 0.05368

t= 0.001/ 0.05368/ √6

t= 0.18629/2.449

t= 0.07607

Since the calculated value of t= 0.07607 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the mean weight differs between the two balances.

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Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

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n_2 = sample of females = 1000

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p_2 = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

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P(-1.96 < \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

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