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Conduct a one-tailed hypothesis test given the information below.
A certain brand of Green Energy light bulbs was advertised as having an average illumination life-span of 2,500 hours. A random sample of 50 bulbs burned out with a mean life-span of 2,470 hours and a sample standard deviation of 140 hours. With a 0.05 level of significance, is the sample mean less than the advertised mean?
A survey of the residents of a new subdivision on why they happened to select that area in which to live. You also wish to secure some information about what they like and do not like about life in the subdivision.
For a given confidence interval, what is the interpretation of a 96% confidence level?
The MacBurger restaurant chain claims that the waiting time of customers for service is normally distributed,
Infer that the difference in the level of satisfaction with service between the two branches.
Determine the highest number of miles (to nearest whole number) that motorist could drive in month and still qualify for discount?
A firm keeps account for 28 customers. 12 of these are defined as new customers and 14 are locally based. If just 3 new customers live locally, how many established customers do not live locally?
There is a statistical deception here. Explain what is deceptive about the bar graph.
Suppose that the mean systolic blood pressure of normal adults is 120 millimeters of mercury (mm Hg) and the standard deviation is 5.6. Assume the variable is normally distributed.
Explain the ANOVA output table values:
Which of following statements would be right concerning a twotail test at 0.05 level of significance? We can conclude that average cottage cheese consumption in America is at least 0.705 pound more or less than 2.75 pounds per person per year.
Assume that the significance level is a= 0.1. Use the given information to find the p-value and the critical value (s). With H1: p = with line through it 1/2 the test statistic is z= - 1.35
Drug sniffing dogs must be 95% accurate. A new dog is being tested and is right in 46 of 50 trials. Find a 95% confidence interval for the proportion of times the dog will be correct.
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