When the population standard deviation is known we perform what is called a what?
We are going to examine two equivalent ways to perform a hypothesis test: the classical approach and the
p-value approach. The classical approach is based on standard deviations. This method compares the test statistic (Z-score) to a critical value (Z-score) from the standard normal table. If the test statistic falls in the rejection zone, you reject the null hypothesis. The p-value approach is based on area under the normal curve. This method compares the area associated with the test statistic to alpha (α), the level of significance (which is also area under the
normal curve). If the p-value is less than alpha, you would reject the null hypothesis. As a past student poetically said: If the p-value is a wee value, Reject Ho Both methods must have: There are four steps required for a hypothesis test: The Classical Method for Testing a Claim about the Population Mean (μ) when the Population Standard Deviation (σ) is KnownExample \(\PageIndex{1}\): A Two-sided Test A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean growth of 1.35 inches/year if a fertilization treatment is applied to the stand. He conducts his experiment, collects data from a sample of 32 plots, and gets a sample mean diameter growth of 1.6 in./year. The population standard deviation for this stand is known to be 0.46 in./year. Does he have enough evidence to support his claim? Solution Step 1) State the null and alternative hypotheses.
Step 2) State the level of significance and the critical value.
Figure 1. The rejection zone for a two-sided test.
Step 3) Compute the test statistic.
$$z = \frac {1.6-1.35}{\frac {0.46}{\sqrt {32}}} =3.07\] Step 4) State a conclusion.
Figure 2. The critical values for a two-sided test when α = 0.05. In this problem, the test statistic falls in the red rejection zone. The test statistic of 3.07 is greater than the critical value of 1.96.We will reject the null hypothesis. We have enough evidence to support the claim that the mean diameter growth is different from (not equal to) 1.35 in./year. Example \(\PageIndex{2}\): A Right-sided Test A researcher believes that there has been an increase in the average farm size in his state since the last study five years ago. The previous study reported a mean size of 450 acres with a population standard deviation (σ) of 167 acres. He samples 45 farms and gets a sample mean of 485.8 acres. Is there enough information to support his claim? Solution Step 1) State the null and alternative hypotheses.
Step 2) State the level of significance and the critical value.
Figure 3. Rejection zone for a right-sided hypothesis test.
Step 3) Compute the test statistic.
$$z = \frac {485.8-450}{\frac {167}{\sqrt {45}}} =1.44\] Step 4) State a conclusion.
Figure 4. The critical value for a right-sided test when α = 0.05.
We fail to reject the null hypothesis. We do not have enough evidence to support the claim that the mean farm size has increased from 450 acres. Example \(\PageIndex{3}\):A Left-sided Test A researcher believes that there has been a reduction in the mean number of hours that college students spend preparing for final exams. A national study stated that students at a 4-year college spend an average of 23 hours preparing for 5 final exams each semester with a population standard deviation of 7.3 hours. The researcher sampled 227 students and found a sample mean study time of 19.6 hours. Does this indicate that the average study time for final exams has decreased? Use a 1% level of significance to test this claim. Solution Step 1) State the null and alternative hypotheses.
Step 2) State the level of significance and the critical value.
Figure 9. The rejection zone for a left-sided hypothesis test.
Step 3) Compute the test statistic.
$$z= \frac {19.6-23}{\frac {7.3}{\sqrt {277}}}\] Step 4) State a conclusion.
Figure 10. The critical value for a left-sided test when α = 0.01.
We reject the null hypothesis. We have sufficient evidence to support the claim that the mean final exam study time has decreased below 23 hours. Testing a Hypothesis using P-valuesThe p-value is the probability of observing our sample mean given that the null hypothesis is true. It is the area under the curve to the left or right of the test statistic. If the probability of observing such a sample mean is very small (less than the level of significance), we would reject the null hypothesis. Computations for the p-value depend on whether it is a one- or two-sided test. Steps for a hypothesis test using p-values:
Instead of comparing Z-score test statistic to Z-score critical value, as in the classical method, we compare area of the test statistic to area of the level of significance. Note:The Decision Rule If the p-value is less than alpha, we reject the null hypothesis. Computing P-valuesIf it is a two-sided test (the alternative claim is ≠), the p-value is equal to two times the probability of the absolute value of the test statistic. If the test is a left-sided test (the alternative claim is “<”), then the p-value is equal to the area to the left of the test statistic. If the test is a right-sided test (the alternative claim is “>”), then the p-value is equal to the area to the right of the test statistic. Let’s look at Example 6 again. A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean growth of 1.35 in./year if a fertilization treatment is applied to the stand. He conducts his experiment, collects data from a sample of 32 plots, and gets a sample mean diameter growth of 1.6 in./year. The population standard deviation for this stand is known to be 0.46 in./year. Does he have enough evidence to support his claim? Step 1) State the null and alternative hypotheses.
Step 2) State the level of significance.
Step 3) Compute the test statistic.
$$z=\frac{1.6-1.35}{\frac{0.46}{\sqrt {32}}}=3.07\] The p-value is two times the area of the absolute value of the test statistic (because the alternative claim is “not equal”). Figure 11. The p-value compared to the level of significance.
Step 4) Compare the p-value to alpha and state a conclusion.
Let’s look at Example 7 again. A researcher believes that there has been an increase in the average farm size in his state since the last study five years ago. The previous study reported a mean size of 450 acres with a population standard deviation (σ) of 167 acres. He samples 45 farms and gets a sample mean of 485.8 acres. Is there enough information to support his claim? Step 1) State the null and alternative hypotheses.
Step 2) State the level of significance.
Step 3) Compute the test statistic.
$$z= \frac {485.8-450}{\frac {167}{\sqrt {45}}}=1.44\] The p-value is the area to the right of the Z-score 1.44 (the hatched area).
Figure 12. The p-value compared to the level of significance for a right-sided test. Step 4) Compare the p-value to alpha and state a conclusion.
We fail to reject the null hypothesis. We do not have enough evidence to support the claim that the mean farm size has increased. Let’s look at Example 8 again. A researcher believes that there has been a reduction in the mean number of hours that college students spend preparing for final exams. A national study stated that students at a 4-year college spend an average of 23 hours preparing for 5 final exams each semester with a population standard deviation of 7.3 hours. The researcher sampled 227 students and found a sample mean study time of 19.6 hours. Does this indicate that the average study time for final exams has decreased? Use a 1% level of significance to test this claim. Step 1) State the null and alternative hypotheses.
Step 2) State the level of significance.
Step 3) Compute the test statistic.
$$z=\frac {19.6-23}{\frac {7.3}{\sqrt {227}}}=-7.02\] The p-value is the area to the left of the test statistic (the little black area to the left of -7.02). The Z-score of -7.02 is not on the standard normal table. The smallest probability on the table is 0.0002. We know that the area for the Z-score -7.02 is smaller than this area (probability). Therefore, the p-value is <0.0002. Figure 13. The p-value compared to the level of significance for a left-sided test. Step 4) Compare the p-value to alpha and state a conclusion.
We reject the null hypothesis. We have enough evidence to support the claim that the mean final exam study time has decreased below 23 hours. Both the classical method and p-value method for testing a hypothesis will arrive at the same conclusion. In the classical method, the critical Z-score is the number on the z-axis that defines the level of significance (α). The test statistic converts the sample mean to units of standard deviation (a Z-score). If the test statistic falls in the rejection zone defined by the critical value, we will reject the null hypothesis. In this approach, two Z-scores, which are numbers on the z-axis, are compared. In the p-value approach, the p-value is the area associated with the test statistic. In this method, we compare α (which is also area under the curve) to the p-value. If the p-value is less than α, we reject the null hypothesis. The p-value is the probability of observing such a sample mean when the null hypothesis is true. If the probability is too small (less than the level of significance), then we believe we have enough statistical evidence to reject the null hypothesis and support the alternative claim. Software SolutionsMinitab(referring to Ex. 8)
One-Sample Z
ExcelExcel does not offer 1-sample hypothesis testing. When the population standard deviation is known we use the?If the population standard deviation (σ) is known, a hypothesis test performed for one population mean is called one-mean z-test or simply z-test.
What is standard deviation known as?Standard Deviation is also known as root-mean square deviation as it is the square root of means of the squared deviations from the arithmetic mean. In financial terms, standard deviation is used -to measure risks involved in an investment instrument.
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