b) A survey of 672 audited tax returns showed that 448 resulted in additional
payments. Construct a 95% confidence interval for true percentage of all audited tax return that resulted in additional payments.
c) A packaging device is set to fill detergent powder packets with a mean weight of 5 kilogram and standard deviation of 0.21 kilogram. The weight of packets can be assumed to be normally distributed. The weight of packet is known to drift upwards over a period of time due to machine fault which is not tolerable. A random sample of 100 packets is taken and weighted. The sample has a mean weight of 5.03 kilogram. Can we conclude that the mean weight produced by machine has increased? Use 5% level of significance.
b) To construct a confidence interval for the true percentage of all audited tax returns that resulted in additional payments, we can use the formula for confidence interval for proportions.
First, we need to compute the sample proportion, which is the ratio of the number of audited tax returns that resulted in additional payments to the total number of audited tax returns:
Sample proportion (p̂) = Number of audited tax returns resulting in additional payments / Total number of audited tax returns = 448/672 = 0.6667
Next, we calculate the standard error of the sample proportion:
Standard Error = sqrt((p̂ * (1-p̂))/n)
where n is the sample size (672 in this case).
Now, we can use the formula for a confidence interval for proportions:
Confidence Interval = p̂ ± (Z * Standard Error)
where Z is the z-score corresponding to the desired confidence level. For a 95% confidence level, the z-score is approximately 1.96.
Plugging in the values:
Confidence Interval = 0.6667 ± (1.96 * sqrt((0.6667 * (1-0.6667))/672))
Simplifying the equation will give you the confidence interval for the true percentage of all audited tax returns that resulted in additional payments.
c) To determine if there is a significant increase in the mean weight produced by the machine, we can perform a hypothesis test.
First, we set up the null hypothesis (H0) and the alternative hypothesis (H1).
H0: The mean weight produced by the machine has not increased (μ = 5)
H1: The mean weight produced by the machine has increased (μ > 5)
Next, we calculate the test statistic, which is the Z-score in this case. The Z-score can be calculated using the formula:
Z = (Sample mean - Population mean) / (Population standard deviation / sqrt(sample size))
In this case, the sample mean is 5.03, the population mean is 5, the population standard deviation is 0.21, and the sample size is 100.
Next, we compare the calculated Z-score to the critical value for the desired level of significance (5% in this case). If the calculated Z-score is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
We can lookup the critical value for a one-tailed test (right-tailed) at the 5% significance level from the standard normal distribution table. For a 5% level of significance, the critical value is approximately 1.645.
If the calculated Z-score is greater than 1.645, we can conclude that the mean weight produced by the machine has increased. Otherwise, we fail to conclude that there is a significant increase.