Answer:
The mean weight is not greater than 16 ounces
Step-by-step explanation:
Null hypothesis: Mean weight is greater than 16 ounces
Alternate hypothesis: Mean weight is not greater than 16 ounces.
Sample mean= 16.04 ounces
Assumed mean= 16 ounces
Significance level = 0.05
Sample standard deviation = 0.08 ounces
Number of samples= 35
To test the claim about the mean, Z = (sample mean - assumed mean) ÷ (standard deviation÷ √number of samples)
Z = (16.04 - 16) ÷ (0.08 ÷ √35)
Z = 0.04 ÷ (0.08 ÷ 5.92)
Z= 0.04 ÷ 0.014 = 2.86
Z = 2.86
It is a one-tailed test, the critical region is an area of 0.05 (the significance level). The critical value of the critical region is 1.64
Since the computed Z 2.86 is greater than the critical value 1.64, the null hypothesis is rejected.
Conclusion
The mean weight is not greater than 16 ounces
<span>500ft/1minute
</span>
<span>The rate which deviates from the choices
given is 500ft/1min
<span>Since, 60mi/hr and 88ft/s =1440mi/day
To illustrate this hypothesis we can solve and convert 60miles/hr
60 x 24 = 1440
hence, 60miles/hr = 1440miles/day
88ft/s
1. 88ft = 0.016667miles
2. 60 seconds = 1 minute </span></span>
3. 60 minutes = 1 hr
4. 24 hours = 1 day
Hence, in conversion
<span><span>
60 sec x 60 min x 24 hours = 86400s
0.016667 miles x 86400 sec = 1440 mi/day
88ft/s = 1440miles/day</span> </span>
The total cost would be 26.50