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Answer:
[tex]z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82[/tex]
[tex]p_v =P(z<-0.82)=0.206[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of newborn boys babies is not significantly lower than 0.508
Step-by-step explanation:
Data given and notation
n=50 represent the random sample taken
[tex]\hat p=0.45[/tex] estimated proportion of newborn boys babies
[tex]p_o=0.508[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95 (asumed)
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We assume that we want to check if the true proportion is less than 0.508.
Null hypothesis:[tex]p\geq 0.508[/tex]
Alternative hypothesis:[tex]p < 0.508[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-0.82)=0.206[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of newborn boys babies is not significantly lower than 0.508
It is concluded that the null hypothesis H₀ is not rejected. There is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.
What is the z test statistic for one sample proportion?
Suppose that we have:
- n = sample size
- [tex]\hat{p}[/tex] = sample proportion
- p₀ = population proportion (hypothesised)
Then, the z test statistic for one sample proportion is:
[tex]Z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}[/tex]
Making the hypothesis and performing the test, assuming that we want to test if the population mean proportion is < 0.508 at the level of significance 0.05.
- (1) Null and Alternative Hypotheses
The following null and alternative hypotheses for the population proportion needs to be tested:
[tex]H_0: p \geq 0.508 \\\\ H_a: p & < & 0.508 \end{array}[/tex]
This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.
- (2) Rejection Region
Based on the information provided, the significance level is [tex]\alpha = 0.05[/tex], and the critical value for a left-tailed test is [tex]z_c = -1.64[/tex]
The rejection region for this left-tailed test is [tex]R = \{z: z < -1.645\}[/tex]
- (3) Test Statistics
The z-statistic is computed as follows:
[tex]z & = & \displaystyle \frac{\hat p - p_0}{\sqrt{ \displaystyle\frac{p_0(1-p_0)}{n}}} \\\\& = & \displaystyle \frac{0.45 - 0.508}{\sqrt{ \displaystyle\frac{ 0.508(1 - 0.508)}{50}}} \\\\ & = & -0.82[/tex]
- (4) Decision about the null hypothesis
Since it is observed that [tex]z = -0.82 \ge z_c = -1.645[/tex], it is then concluded that the null hypothesis is not rejected.
Thus, it is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.
Learn more about population proportion here:
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