Apply the 6 steps in hypothesis testing:1. Formulate the H0 and Ha2. Choose test statistic (such as z- test, t-test, or F-test)(? unknown = t-test; ? known = z- test)3. Determine the critical region/ rejection region4. Calculate the test statistic5. Provide statistical decision (Reject/ or not to reject H0)6. Make an interpretation and if possible, draw a conclusion.A pediatrician claimed that the average time for a drug to take effect is 15 minutes with a standard deviation of eight minutes. In a sample of 49 trials, the average time was 18 minutes. Test the claim of a pediatrician that the average time is equal to 15 minutes. Assume that the average time for a drug to take effect is approximately normally distributed. In 2006, the average annual family income in the National Capital Region was P310,860 with standarddeviation of P176,870. With the increased prices of commodities, it is hoped that the mean annualfamily income has also increased. A survey is conducted to verify this contention. A random sample of320 families has a mean annual income of P325,255. Isthe sample result sufficient to demonstrate thatthe mean annual family income in NCR is significantly higher than P310,860 at 1% level of significance?1. H0: µ=310,860 OR HO: µ<310,860Ha: µ>310,860 (claim)2. Since o is known the test statistic is z – test3. The level of significance is a = 0.01 Type of test: one- tailed (right)cv=2.334. µ=310,860 x=325,255 0=176,870 n=320z=325255-310860 =1.46176870V3205. Since |zcomp|=1.46<|zcrit|=2.33, then Do Not Reject the HO6. There is a statistical evidence to show that the mean annual family income in NCR is not higherthan P310,860 at 1% level of significance.

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Description: Apply the 6 steps in hypothesis testing:1. Formulate the H0 and Ha2. Choose test statistic (such as z- test, t-test, or F-test)(? unknown = t-test; ? known = z- test)3. Determine the critical region/ rejection region4. Calculate the test statistic5.

1. Let μ denotes the population mean time for a drug. The claim of a pediatrician that the average…

Answered: A pediatrician claimed that the average…

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Apply the 6 steps in hypothesis testing:

1. Formulate the H0 and Ha
2. Choose test statistic (such as z- test, t-test, or F-test)(? unknown = t-test; ? known = z- test)
3. Determine the critical region/ rejection region
4. Calculate the test statistic
5. Provide statistical decision (Reject/ or not to reject H0)
6. Make an interpretation and if possible, draw a conclusion.

A pediatrician claimed that the average time for a drug to take effect is 15 minutes with a standard deviation of eight minutes. In a sample of 49 trials, the average time was 18 minutes. Test the claim of a pediatrician that the average time is equal to 15 minutes. Assume that the average time for a drug to take effect is approximately normally distributed.

In 2006, the average annual family income in the National Capital Region was P310,860 with standard
deviation of P176,870. With the increased prices of commodities, it is hoped that the mean annual
family income has also increased. A survey is conducted to verify this contention. A random sample of
320 families has a mean annual income of P325,255. Isthe sample result sufficient to demonstrate that
the mean annual family income in NCR is significantly higher than P310,860 at 1% level of significance?
1. H0: µ=310,860 OR HO: µ<310,860
Ha: µ>310,860 (claim)
2. Since o is known the test statistic is z – test
3. The level of significance is a = 0.01 Type of test: one- tailed (right)
cv=2.33
4. µ=310,860 x=325,255 0=176,870 n=320
z=325255-310860 =1.46
176870
V320
5. Since |zcomp|=1.46<|zcrit|=2.33, then Do Not Reject the HO
6. There is a statistical evidence to show that the mean annual family income in NCR is not higher
than P310,860 at 1% level of significance.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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