Hypothesis and level of significance
Null hypothesis
Ho: µ = M
The null hypothesis shows that the population mean (µ) is equal to the sample mean (M). In this case, the value of the sample mean is 94. The null hypothesis implies that the children’s intelligence is not affected if their mothers have German measles during pregnancy. It could also mean that the German Measles does not affect the intelligence quotient of a child.
Alternative hypothesis
H1: µ ≠ M
The alternative hypothesis shows that the population means is not equal to the sample mean. This implies that a child’s intelligence is affected if their mothers have German measles during pregnancy. It could also mean that the German measles affects the intelligence quotient of a child.
It can be noted that the null and the alternative hypotheses are mutually exclusive. This relationship is important, especially when making decisions. The test will be carried out at the 95% confidence level. The confidence level gives the proportion of all the samples that can contain the decision criterion. For instance, if the confidence level is 95% like in this case, then it implies the proportion of the sample that can contain the population variable. Therefore, the confidence level is important when deciding on whether to reject or not reject the null hypothesis. Thus, the significance level or alpha is 0.05. This is derived from the p-value that is provided.
Statistical test
In order to test the hypothesis stated above, it is important to select a suitable statistical test. There are a number of statistical tests that can be used to test the hypothesis. Examples are the F test, t-test, the Z test, and Chi – square test among others. In this case, only two tests will be relevant. These are Z test and t test. The Z-test is used when the size of the sample is large or when the standard deviation of the population is known. On the other hand, t-test is used when the sample size is small, less than thirty, or if the population variance is not known. From the data provided, it can be noted that the population mean and standard deviation are known. Besides, the sample size is small, less than thirty. Therefore, a Z-test will be used to test the hypothesis. If the population variance was not stated, then the t-test would have been more suitable. The two tests require the population to be normally distributed. The hypothesis stated above requires that a two tailed test to be carried. A two tailed test implies that the decision on whether to reject or not is based on two extreme values unlike the case of one tailed test where the decision is based on one extreme value, either on the right or on the left depending on the sign of the test statistic. Therefore, a two tailed one sample z-test will be carried out to test the hypothesis.
Values that will be compared
In this case, the values that will be compared are the z-statistic (that is the value of z that is calculated) and z critical (this value is obtained from the table).
Decision rule
The null hypothesis will not be rejected if the value of z statistic is lower than the value of z critical. On the other hand, the null hypothesis will be rejected and lean on the alternative hypothesis if the value of of z statistic is greater than the value of z critical.
Directional versus non directional alternative hypothesis
A directional alternative hypothesis serves two purposes. First, it negates the null hypothesis by stating that it is wrong. Secondly, it indicates whether the value of the variable being analyzed is more than or less than the reference value. On the other hand, a non directional alternative hypothesis serves only one purpose, that is, it only negates the alternative hypothesis. It does not compare the magnitude of the parameter being examined. A directional alternative hypothesis is useful because it makes the analysis more comprehensive. However, it lacks the ability to detect the outcomes that occur in the reverse direction. In the example above, it can be noted that the alternative hypothesis is non directional because the aim of the test it to establish whether or not that child’s intelligence is affected if their mothers have German measles during pregnancy. Therefore, the response can take either side. In most cases, a two tailed test will lead to a non directional alternative hypothesis while a one tailed test leads to a directional alternative hypothesis.
Calculations
In this section value of z calculated and z critical will be estimated. The value of z will be computed using the formula shown below.
- Z = (µ – M) / (δ / √n)
These variables are:
- µ – population mean (this value is 100)
- M – sample mean (this value is 94)
- δ – population standard deviation (this value is 15)
- n – sample size (the value is 25)
Substitute the values in the formula above to obtain z statistic:
Z = (100 – 94) / (15 / √25)
= 6 / (15 / 5)
= 6 / 3
= 2
After computing the z statistics, then it is important to estimate the value of z critical. This value will be obtained from the standard normal table. The significance level is 5%. Since the hypothesis will be tested using a two tailed test, the 5% will be split into two. Thus, 0.025 will represent the lower tail. To obtain the upper tail, 0.025 will be deducted from 1. Thus, the upper tail will be 0.975. Therefore, the z value that matches to 0.025 is 1.96. Therefore, the value that corresponds to lower tail will be -1.96 while the upper tail will be 1.96. Therefore, the rejection region will be outside the region between -1.96 and 1.96
Conclusion
From the discussion above, the null hypothesis stated that a child’s intelligence is not affected if their mothers have German measles during pregnancy while the alternative hypothesis child’s intelligence is affected if their mothers have German measles during pregnancy. Further, a two tailed z-test was carried out to test this hypothesis. The test was carried out at the 95% confidence level. In the calculations presented in step three, it can be observed that the value of z statistic is greater than the value z critical. This implies that the z statistic falls in the rejection region. Therefore, the null hypothesis will be rejected at the 95% confidence level. This implies that a child’s intelligence is interfered with if their mothers have German measles during pregnancy. In conclusion, the data used to carry out the test reveal that German measles have a significant impact on intelligence quotient of a child at 5% significance level.