A food distribution company claims that a fast food restaurant chain receives, on average, 24 pounds of meat on a daily

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A food distribution company claims that a fast food restaurant chain receives, on average, 24 pounds of meat on a daily

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A Food Distribution Company Claims That A Fast Food Restaurant Chain Receives On Average 24 Pounds Of Meat On A Daily 1
A Food Distribution Company Claims That A Fast Food Restaurant Chain Receives On Average 24 Pounds Of Meat On A Daily 1 (59.8 KiB) Viewed 28 times
A Food Distribution Company Claims That A Fast Food Restaurant Chain Receives On Average 24 Pounds Of Meat On A Daily 2
A Food Distribution Company Claims That A Fast Food Restaurant Chain Receives On Average 24 Pounds Of Meat On A Daily 2 (41.19 KiB) Viewed 28 times
A food distribution company claims that a fast food restaurant chain receives, on average, 24 pounds of meat on a daily basis. The district manager of the restaurant chain decides to randomly sample 31 shipments from the company and finds a mean weight of 22.5 pounds with a standard deviation of 4.4 pounds. Test at a 3% level of significance to determine whether or not the food distribution company sends less than they claim. a. Check all of the requirements that are satisfied. random the i distribution is normal since n > 30 the 7 distribution is normal since the distribution is normal the p distribution is normal since np > 10 and nq 2 10 b. Identify the null and alternative hypotheses. H: OM H: C. What type of hypothesis test should you conduct (left, right, or two-tailed)? left-tailed right-tailed two-tailed d. Identify the appropriate significance level. Make sure to enter your answer as a decimal. e. Which calculator function should you use? T-Test f. Find the test statistic. Write the result below, and be sure to round your final answer to two decimal places. g. Find the p-value. Enter your answer as a decimal (not a percentage) and round to 4 decimal places.

Test the claim that the proportion of people who own cats is larger than 50% at the 0.05 significance level. The null and alternative hypothesis would be: Họ:1 = 0.5 Hp:u < 0.5 Họ:> 0.5 Ho:p> 0.5 Ho:p < 0.5 Ho:p = 0.5 H:14 0.5 Hậu > 0.5 H:1 < 0.5 Hip < 0.5 Hip > 0.5 Hộp + 0.5 The test is: right-tailed two-tailed left-tailed 0 Based on a sample of 100 people, 58% owned cats The p-value is: (to 2 decimals) Based on this we: Fail to reject the null hypothesis O Reject the null hypothesis Question Help: D Video
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