Hello, I'm looking for a solution to this activities. The activities is given in the google drive link below, and any further concerns will be addressed in the comments area. I hope you can assist me, and I will undoubtedly provide you with 100% positive feedback.
https://drive.google.com/drive/folders/1jtlLuhFwULjYQxBoVUO60XIVhCU628VD?usp=sharing
FOR REFERENCES:
O a Let Us Study There are two specic tests statistics used for hypothesis testing concerning means: ztest and ttest. When to use? when a is known and n 2 30 or the z-test where _ population is x 2 sample mean normally distributed H = population mean ifn |criticalvalue| not supported Table 3. Rejection Region Approach If the absolute value of the computed zvalue 2c is greater than the absolute value of the critical value, we reject the null hypothesis and support the alternative hypothesis. But if the absolute value of the computed value is less than the absolute value of the critical value, we do not reject or we fail to reject the null hypothesis and the alternative hypothesis is not supported. In a right-tailed test, if the computed value is greater than the critical value, we reject the null hypothesis and support the alternative hypothesis. But if the computed value is less than the critical value, we do not reject or we fail to reject the null hypothesis and the alternative hypothesis is not supported. In a left-tailed test, if the computed value is less than the critical value, we reject the null hypothesis and support the alternative hypothesis. But if the computed value is greater than the critical value, we do not reject or we fail to reject the null hypothesis and the alternative hypothesis is not supported. Rejecting the null hypothesis doesn't mean that it is incorrect or the alternative hypothesis is correct. The collected data suggest a sufcient evidence to disprove the null hypothesis, hence we reject it. Similarly, a failure to reject the null hypothesis does not mean that it is true only that the test did not prove it to be false. There is an insufcient evidence to disprove the null hypothesis; hence we do not reject it. Study the given examples below. Example 1: According to a study conducted by the Grade 12 students, P155 is the average monthly expense for cell phone loads of senior high school students in their province. A Statistics student claims that this amount has increased since January of this year. Do you think his claim is acceptable if a random sample of 50 students has an average monthly expense of ?165 for cell phone loads? Using 5% level of signicance, assume that a population standard deviation is P52. Solution: [1:155 0:52 a=0.05 Gwen: a? = 165 n = 50 Step 1: State the Null Hypothesis: null and alternative The average monthly expense for cellphone loads is hypotheses. ?155. Halli = 155 Alternative Hypothesis: The average monthly expense for cellphone loads is more than F155. Ham > 155 Step 2: Determine Since the population mean is being tested, the the test statistic, population standard deviation 0 is known, and n > then compute its test 30, the appropriate test statistic is the z-test. value. _ ru _ 165155 _ 10 2 l _ 52 _ 735 W E ' z = 1. 361 (test statistic or z computed value) Step 3: Find the critical value using the zCritical Value Table Table 1: z Critical Value Type of Test Onetailed Test TWOtailed Test c1 =.05 or 5% Zc = H.645 Z: = i1.960 Since a =0.05 and the alternative hypothesis Ha: p > 155 is directional (righttailed test), the critical value is z = 1.645. Step 4: Draw the . . curve, show the Non-Rejection ' ' _ Region _ . ' rejection and non Rejectlon Region rejection regions, then make a decision. 1.361I'E45 *Recall that the From Table 1, critical region _ . , , ,, . corresponds to the Right-tailed lteststahstlclz|cr1t1calvalue| Reject Ho and sugportll rejection of the null test lteststatisticl then compute its test value. 30, the appropriate test statistic is the z-test. T-u_95-99_ 4 Z: 1 'T m x/ m ' z = 1. 688 (test statistic or z computed value) Step 3: Find the critical value using the zCritical Value Table Table 1: z Critical Value Type of Test Twotailed Test (1 =.05 or 5% Since a =0.05 and the alternative hypothesis Ha: u |critical valuel Since the test statistic or zcomputed value, 1.688 it lies within the rejection region, we reject the null hypothesis. Step 5: Summarize There is enough evidence to support the claim that the results. the average IQ level of Senior High School students is lower than 99. This result is significant at a = 0.05 level

Question

Hello, I'm looking for a solution to this activities. The activities is given in the google drive link below, and any further concerns will be addressed in the comments area. I hope you can assist me, and I will undoubtedly provide you with 100% positive feedback.

https://drive.google.com/drive/folders/1jtlLuhFwULjYQxBoVUO60XIVhCU628VD?usp=sharing

FOR REFERENCES:

O a Let Us Study There are two specic tests statistics used for hypothesis testing concerning means: ztest and ttest. When to use? when a is known and n 2 30 or the z-test where _ population is x 2 sample mean normally distributed H = population mean ifn |criticalvalue| not supported Table 3. Rejection Region Approach If the absolute value of the computed zvalue 2c is greater than the absolute value of the critical value, we reject the null hypothesis and support the alternative hypothesis. But if the absolute value of the computed value is less than the absolute value of the critical value, we do not reject or we fail to reject the null hypothesis and the alternative hypothesis is not supported. In a right-tailed test, if the computed value is greater than the critical value, we reject the null hypothesis and support the alternative hypothesis. But if the computed value is less than the critical value, we do not reject or we fail to reject the null hypothesis and the alternative hypothesis is not supported. In a left-tailed test, if the computed value is less than the critical value, we reject the null hypothesis and support the alternative hypothesis. But if the computed value is greater than the critical value, we do not reject or we fail to reject the null hypothesis and the alternative hypothesis is not supported. Rejecting the null hypothesis doesn't mean that it is incorrect or the alternative hypothesis is correct. The collected data suggest a sufcient evidence to disprove the null hypothesis, hence we reject it. Similarly, a failure to reject the null hypothesis does not mean that it is true only that the test did not prove it to be false. There is an insufcient evidence to disprove the null hypothesis; hence we do not reject it. Study the given examples below. Example 1: According to a study conducted by the Grade 12 students, P155 is the average monthly expense for cell phone loads of senior high school students in their province. A Statistics student claims that this amount has increased since January of this year. Do you think his claim is acceptable if a random sample of 50 students has an average monthly expense of ?165 for cell phone loads? Using 5% level of signicance, assume that a population standard deviation is P52. Solution: [1:155 0:52 a=0.05 Gwen: a? = 165 n = 50 Step 1: State the Null Hypothesis: null and alternative The average monthly expense for cellphone loads is hypotheses. ?155. Halli = 155 Alternative Hypothesis: The average monthly expense for cellphone loads is more than F155. Ham > 155 Step 2: Determine Since the population mean is being tested, the the test statistic, population standard deviation 0 is known, and n > then compute its test 30, the appropriate test statistic is the z-test. value. _ ru _ 165155 _ 10 2 l _ 52 _ 735 W E ' z = 1. 361 (test statistic or z computed value) Step 3: Find the critical value using the zCritical Value Table Table 1: z Critical Value Type of Test Onetailed Test TWOtailed Test c1 =.05 or 5% Zc = H.645 Z: = i1.960 Since a =0.05 and the alternative hypothesis Ha: p > 155 is directional (righttailed test), the critical value is z = 1.645. Step 4: Draw the . . curve, show the Non-Rejection ' ' _ Region _ . ' rejection and non Rejectlon Region rejection regions, then make a decision. 1.361I'E45 *Recall that the From Table 1, critical region _ . , , ,, . corresponds to the Right-tailed lteststahstlclz|cr1t1calvalue| Reject Ho and sugportll rejection of the null test lteststatisticl then compute its test value. 30, the appropriate test statistic is the z-test. T-u_95-99_ 4 Z: 1 'T m x/ m ' z = 1. 688 (test statistic or z computed value) Step 3: Find the critical value using the zCritical Value Table Table 1: z Critical Value Type of Test Twotailed Test (1 =.05 or 5% Since a =0.05 and the alternative hypothesis Ha: u |critical valuel Since the test statistic or zcomputed value, 1.688 it lies within the rejection region, we reject the null hypothesis. Step 5: Summarize There is enough evidence to support the claim that the results. the average IQ level of Senior High School students is lower than 99. This result is significant at a = 0.05 level