Imagine you are a manager at a major bottling company. Customers have begun to complain that the bottles of the brand of soda produced in your company contain less than the advertised sixteen (16) ounces of product. Your boss wants to solve the problem at hand and has asked you to investigate. You have your employees pull thirty (30) bottles off the line at random from all the shifts at the bottling plant. You ask your employees to measure the amount of soda there is in each bottle. Bottle Number | Ounces | Bottle Number | Ounces | Bottle Number | Ounces | 1 | 14.5 | 11 | 15 | 21 | 14.1 | 2 | 14.6 | 12 | 15.1 | 22 | 14.2 | 3 | 14.7 | 13 | 15 | 23 | 14 | 4 | 14.8 | 14 | 14.4 | 24 | 14.9 | 5 | 14.9 | 15 | 15.8 | 25 | 14.7 | 6 …show more content…
The customers in this case study have complained that the bottling company provides less than the advertised sixteen ounces of product. They need to determine if there is enough evidence to conclude the soda bottles do not contain sixteen ounces. The sample size of sodas is 30 and has a mean of 14.9. The standard deviation is found to be 0.55. With these calculations and a confidence level of 95%, the confidence interval would be 0.2. There is a 95% certainty that the true population mean falls within the range of 14.7 to 15.1. Conduct a Hypothesis Test to Verify if Claim is Supported. “Hypothesis testing is a decision-making process for evaluating claims about a population” (Bluman, 2013, p. 398). This process is used to determine if you will accept or reject the hypothesis. The claim is that the bottles contain less than 16 ounces. The null hypothesis is the soda bottles contain 16 ounces. The alternative hypothesis is the bottles contain less than 16 ounces. The significance level will be 0.05. The test method to be used is a t-score. The test statistic is calculated to be -11.24666539 and the P-value is 1.0. The P-value is the probability of observing a sample statistic as extreme as the test statistic, assuming the null hypothesis is true. The T Crit value is 1.69912702. The calculations show there is enough evidence to support the claim that the soda bottles do
Testing allows the p-value that represents the probability showing that results are unlikely to occur by chance. A p-value of 5% or lower is statistically significant. The p value helps in minimizing Type I or Type II errors in the dataset that can often occur when the p value is more than the significance level. The p value can help in stopping the positive and negative correlation between the dataset to reject the null hypothesis and to determine if there is statistical significance in the hypothesis. Understanding the p value is very important in helping researchers to determine the significance of the effect of their experiment and variables for other researchers
12. _____ For a given population, confidence intervals constructed from larger samples tend to be narrower than those constructed from smaller samples. Which statement below best describes why this is true? (A) The variability of the sample mean is less for larger samples. (B) The z-value for larger samples tends to be more accurate. (C) The population variance is larger for large populations. (D) As the sample size increases, the z-value (or t-value) becomes smaller. A machine dispenses potato chips into bags that are advertised as containing one pound of product. To be on the safe side, the machine is supposed to be calibrated to dispense 16.07 ounces per bag, and from long time observation, the distribution of the fill-weights is known to be approximately normal and the process is known to have a standard deviation of 0.15 ounces.
Consider the following scenario in answering questions 5 through 7. In an article appearing in Today’s Health a writer states that the average number of calories in a serving of popcorn is 75. To determine if the average number of calories in a serving of popcorn is different from 75, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 78 with a sample standard deviation of 7.
Since the P-value (0.386) is greater than the significance level (0.05), we fail to reject the null hypothesis. The p-value implies the probability of rejecting a true null hypothesis.
The null hypothesis was that the female and male shoe sizes have an equal mean while the alternative hypothesis was that female and male shoe sizes do not have an equal mean. With the degrees of freedom being 33, the t-statistic is -8.27. The probability that -8.27 is ≤-1.69 is 7.5×10-10 for the one-tailed test. Also, the probability that -8.27 is ≤ ±2.03. is 1.5×10-9 for the two-tailed test. Due to both probabilities being under the alpha value of 0.05, the null hypothesis is rejected, and the alternative hypothesis is accepted at the 95% confidence level.
A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true proportion of families who own at least on DVD player. Place your limits, rounded to 3 decimal places, in the blanks. Place the lower limit in the first blank and the upper limit in the second blank When entering your answer do not use any labels or symbols other than the decimal point. Simply provide the numerical values. For example, 0.123 would be a legitimate entry.
46. As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 220 applicants 14 failed the test. Develop a 99 percent confidence interval for the proportion of applicants that fail the test. Would it be reasonable to conclude that more than 10 percent of the applicants are now failing the test? In addition to the testing of applicants, Fashion Industries randomly tests its employees throughout the year. Last year in the 400 random tests conducted, 14 employees failed the test. Would it be reasonable to conclude that less than 5 percent of the employees are not able to pass the random drug test?
It tells that the t-statistic with 97 degrees of freedom was 2.14, and the corresponding p-value was less than .05, specifically around 0.035. Therefore, it is appropriate to conclude the research study was statistically significant.
1. A researcher is interested in whether students who attend private high schools have higher average SAT Scores than students in the general population. A random sample of 90 students at a private high school is tested and and a mean SAT score of 1030 is obtained. The average score for public high school student is 1000 (σ= 200).
To test the null hypothesis, if the P-Value of the test is less than 0.05 I will reject the null hypothesis.
Provide the following discussion based on the conclusion of your test: If you conclude that there are less than sixteen (16) ounces in a bottle of soda, speculate on three (3) possible causes. Next, suggest the strategies to avoid the deficit in the future.
Test statistic t = -3.937 (df = 80), p-value < .005, reject Hₒ and conclude that there is a significant
A cereal manufacturer uses a filling process designed to add exactly 18 ounces of cereal to each box. State the null and alternative hypotheses that would be used to verify this claim.
(14.87*16/0.550329055/5.4772 = 0.1005) So, we have p = 0.1005. Then we ask if p is greater than a (alpha). Since p is greater than a = 0.10 we decide that there is significant evidence supporting the claim to not reject the Ho. So, we accept the null hypothesis that on average there is less than the advertised 16oz in each bottle.
The confidence interval for proportion of business students of bayview University who cheat is some form is 0.43 to 0.63.