Question

please answer the following problems numerically

2. Does an average box of cereal more than 368 grams of cereal? A random sample of 49 boxes showed sample mean = 360. The company has specified Ito be 15 grams and the distribution to be normal. Test at the IM00.05 level. 3. Does an average box of cereal less than 378 grams of cereal? A random sample of 64 boxes showed sample mean = 372. The company has specified sample standard deviationſ|to be 15 grams and the distribution to be normal. Test at the 0000.10 4. Does an average box of cereal less than 378 grams of cereal? A random sample of 64 boxes showed mean = 370. The company has specified to be 15 grams and the distribution to be normal. O7000.10 5. Does an average box of cereal less than 378 grams of cereal? A random sample of 64 boxes showed mean = 370. The company has specified to be 15 grams and the distribution to be normal. 00000.10

Answer 1

The provided sample mean is X = 360 and the known population standard deviation is o = 15, and the sample size is n = 49. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 368 Ha: ji > 368 This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a right-tailed test is ze = 1.64. The rejection region for this right-tailed test is R={z: z > 1.64} st Statistics The z-statistic is computed as follows: Z = X – Mo olvn = 360 – 368 -= -3.733 15/49 (4) Decision about the null hypothesis Since it is observed that z = -3.733 < ze = 1.64, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.9999, and since p= 0.9999 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is greater than 368, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 355.8 < d < 364.2.

The provided sample mean is X = 372 and the known population standard deviation is o = 15, and the sample size is n = 64. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 378 Ha: j < 378 This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.10, and the critical value for a left-tailed test is ze = -1.28. The rejection region for this left-tailed test is R= {z:z < -1.28} (3) Test Statistics The z-statistic is computed as follows: Ã - Mo g/ 372 – 378 15/464 = -3.2 15/16 (4) Decision about the null hypothesis Since it is observed that z = -3.2 < Zc = -1.28, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0007, and since p = 0.0007 < 0.10, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is less than 378, at the 0.10 significance level. Confidence Interval The 90% confidence interval is 368.916 < d < 375.084.

The provided sample mean is Ž = 370 and the known population standard deviation is o = 15, and the sample size is n = 64. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p = 378 Ha: j < 378 This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.10, and the critical value for a left-tailed test is ze = -1.28. The rejection region for this left-tailed test is R= {z:z < -1.28} (3) Test Statistics The z-statistic is computed as follows: z =- Ã - Mo olvn 370 – 378 15/164 = -4.267 (4) Decision about the null hypothesis Since it is observed that z = -4.267 < zc = -1.28, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0, and since p=0 < 0.10, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is less than 378, at the 0.10 significance level. Confidence Interval The 90% confidence interval is 366.916 < u < 373.084.

The provided sample mean is X = 370 and the known population standard deviation is o = 15, and the sample size is n = 64. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: ji = 378 Ha: u < 378 This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.10, and the critical value for a left-tailed test is ze = -1.28. The rejection region for this left-tailed test is R= {z:z < -1.28} (3) Test Statistics The Z-statistic is computed as follows: X - Mo olvn 370 – 378 15/764 = -4.267 (4) Decision about the null hypothesis Since it is observed that z = -4.267 < 2c = -1.28, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0, and since p = 0) < 0.10, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is less than 378, at the 0.10 significance level. Confidence Interval The 90% confidence interval is 366.916 < u < 373.084.
its normal distribution problem