Answer:
(a) The sample sizes are 6787.
(b) The sample sizes are 6666.
Step-by-step explanation:
(a)
The information provided is:
Confidence level = 98%
MOE = 0.02
n₁ = n₂ = n
Compute the sample sizes as follows:
Thus, the sample sizes are 6787.
(b)
Now it is provided that:
Compute the sample size as follows:
![n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%28z_%7B%5Calpha%2F2%7D%29%5E%7B2%7D%5Ctimes%20%5B%5Chat%20p_%7B1%7D%281-%5Chat%20p_%7B1%7D%29%2B%5Chat%20p_%7B2%7D%281-%5Chat%20p_%7B2%7D%29%5D%7D%7BMOE%5E%7B2%7D%7D)
![=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2.33%5E%7B2%7D%5Ctimes%20%5B0.45%281-0.45%29%2B0.58%281-0.58%29%5D%7D%7B0.02%5E%7B2%7D%7D%5C%5C%5C%5C%3D6665.331975%5C%5C%5C%5C%5Capprox%206666)
Thus, the sample sizes are 6666.
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.
Answer:
D
Step-by-step explanation:
