7.The Daytona Beach Tourism Commission recently claimed that the average amount of money a typical college student spends per day during spring break is over $70. Based upon previous research, the population standard deviation is estimated to be $17.32. The Commission surveys 35 students and find that the mean spending is $67.57. Is there evidence that the average amount spent by students is less than $70? Use 95% confidence level . Test : : 70 : < 70 Note : Just write the decision (Accept or Reject ) and why do you accept or reject ? Answer :
8. The manufacturer of a certain chewing gum claims that four out of five dentists surveyed prefer their type of gum. You decide to test their claim. You find that in a sample of 200 doctors, 74.1% do actually prefer their gum. Is this evidence sufficient to doubt the manufacturer’s claim? Use = 0.025 Test: : P = 0.80 vs.: P < 0.80 Note : Just write the decision (Accept or Reject ) and why do you accept or reject ? Answer :
Mohmmad Jamal hi! First, please note that my video is on t-test. The problem you posted is on z-test as the pipulation stndard deviation is known. Anyway, perform a z test with null hypothesis that the average amount of money is at least 70(some would set equal to 70 which is acceptable) and alternative hypothesis as the average amount of money is less than 70. Note that if you change how you select your null hypothesis and alternative hypothesis, the succeeding steps will vary. But let us say that those are the null and the alternative hypotheses set. When you perform z- test, you will end up with a conclusion “Do not reject the null hypothesis.” Reason, the z value computed is -0.83 which is outside the critical region as the boundary is at -1.645( z of 0.05). This means, there is no sufficient evidence supporting that the amount of money is less than 70 and that the amound of money should be at least 70(null hypothesis). Also, the problem states the confidence level instead of the significance level. So I am not sure if you are expected to create a confidence intervals instead. But for hypothesis testing, that is my answer.
For number 8 question, I need to know if 0.025 is alpha or half of alpha. Not knowing it exactly will affect the correctness of my conclusion. Also, were you the one who set the null and alternative hypothesis? If your null hypothesis is p= 0.80, the alternative hypothesis must be “p is NOT EQUAL TO 0.80”. I need this info correct before I can give conclusions. Now if the alternative hypothesis is really “p is less than 0.80”, this will be the conclusion. Reject null hypothesis as the computed z will be -2.09 which will lie on the critical region with boundary of -1.96. Therefore, there is sufficient evidence to say the p is less than 0.80. Again, this is the conclusion based on the assumption that the alternative hypothesis is p is less than 0.80. Thank you.
7.The Daytona Beach Tourism Commission recently claimed that the average amount of money a typical college student spends per day during spring break is over $70. Based upon previous research, the population standard deviation is estimated to be $17.32. The Commission surveys 35 students and find that the mean spending is $67.57. Is there evidence that the average amount spent by students is less than $70? Use 95% confidence level .
Test :
: 70
: < 70
Note : Just write the decision (Accept or Reject ) and why do you accept or reject ?
Answer :
8. The manufacturer of a certain chewing gum claims that four out of five dentists surveyed prefer their type of gum. You decide to test their claim. You find that in a sample of 200 doctors, 74.1% do actually prefer their gum. Is this evidence sufficient to doubt the manufacturer’s claim? Use = 0.025
Test:
: P = 0.80 vs.: P < 0.80
Note : Just write the decision (Accept or Reject ) and why do you accept or reject ?
Answer :
Mohmmad Jamal hi! First, please note that my video is on t-test. The problem you posted is on z-test as the pipulation stndard deviation is known. Anyway, perform a z test with null hypothesis that the average amount of money is at least 70(some would set equal to 70 which is acceptable) and alternative hypothesis as the average amount of money is less than 70. Note that if you change how you select your null hypothesis and alternative hypothesis, the succeeding steps will vary. But let us say that those are the null and the alternative hypotheses set. When you perform z- test, you will end up with a conclusion “Do not reject the null hypothesis.” Reason, the z value computed is -0.83 which is outside the critical region as the boundary is at -1.645( z of 0.05). This means, there is no sufficient evidence supporting that the amount of money is less than 70 and that the amound of money should be at least 70(null hypothesis). Also, the problem states the confidence level instead of the significance level. So I am not sure if you are expected to create a confidence intervals instead. But for hypothesis testing, that is my answer.
Pls help me 😭
For number 8 question, I need to know if 0.025 is alpha or half of alpha. Not knowing it exactly will affect the correctness of my conclusion. Also, were you the one who set the null and alternative hypothesis? If your null hypothesis is p= 0.80, the alternative hypothesis must be “p is NOT EQUAL TO 0.80”. I need this info correct before I can give conclusions.
Now if the alternative hypothesis is really “p is less than 0.80”, this will be the conclusion. Reject null hypothesis as the computed z will be -2.09 which will lie on the critical region with boundary of -1.96. Therefore, there is sufficient evidence to say the p is less than 0.80. Again, this is the conclusion based on the assumption that the alternative hypothesis is p is less than 0.80. Thank you.
Watch my video on hypothesis testing on one population proportion.ua-cam.com/video/SJc58kDjH1M/v-deo.html