A law firm is going to designate associates and partners to a big new case. The
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The confidence interval is from 9.81 to 10.19.
We first find the mean of the data:
<span>(9.8+10.2+10.4+9.8+10.0+10.2+9.6)/7 = 10
Next we find the standard deviation:
</span>σ=√([<span>(9.8-10)^2+(10.2-10)^2+(10.4-10)^2+(9.8-10)^2+(10-10)^2+(10.2-10)^2+(9.6-10)^2]/7) = 0.262
The z-score for 95% confidence is found by
1-0.95 = 0.05; 0.05/2 = 0.025; from the z-table, it is 1.96.
The confidence interval is calculated using
</span>
<span>
</span>
We first find the mean of the data:
<span>(9.8+10.2+10.4+9.8+10.0+10.2+9.6)/7 = 10
Next we find the standard deviation:
</span>σ=√([<span>(9.8-10)^2+(10.2-10)^2+(10.4-10)^2+(9.8-10)^2+(10-10)^2+(10.2-10)^2+(9.6-10)^2]/7) = 0.262
The z-score for 95% confidence is found by
1-0.95 = 0.05; 0.05/2 = 0.025; from the z-table, it is 1.96.
The confidence interval is calculated using
</span>
</span>
Answer:
There is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 300
p = 5% = 0.05
Alpha, α = 0.05
Number of dead pixels , x = 24
First, we design the null and the alternate hypothesis
This is a one-tailed(right) test.
Formula:
Putting the values, we get,
Now, we calculate the p-value from excel.
P-value = 0.00856
Since the p-value is smaller than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Conclusion:
Thus, there is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.