Answer:
a
The null hypothesis is
[tex]H_o : \mu =[/tex]$98
The alternative hypothesis is
[tex]H_a : \mu >[/tex]$98
b
test statistics [tex]t_s = 1.091[/tex]
c
The the result of the test statistics is significant
Step-by-step explanation:
From the question we are told that
The population mean is [tex]\mu =[/tex]$98
The standard deviation is [tex]\sigma =[/tex]$11
The sample size is [tex]n = 36[/tex]
The sample mean is [tex]\= x =[/tex]$100
The level of significance is [tex]\alpha = 5[/tex]% = 0.05
The null hypothesis is
[tex]H_o : \mu =[/tex]$98
The alternative hypothesis is
[tex]H_a : \mu >[/tex]$98
Now the critical values for this level of significance obtained from critical value for z-value table is [tex]z_\alpha = 1.645[/tex]
The test statistics is mathematically evaluated as
[tex]t_s = \frac{\= x - \mu}{ \frac{\sigma }{ \sqrt{n} } }[/tex]
substituting values
[tex]t_s = \frac{100 - 98}{ \frac{11 }{ \sqrt{36} } }[/tex]
[tex]t_s = 1.091[/tex]
Looking at [tex]z_\alpha \ and \ t_s[/tex] we see that [tex]z_\alpha \ > t_s[/tex] hence the we fail to reject the null hypothesis
hence there is no sufficient evidence to conclude that the mean weekly food budget for all households in this community is higher than the national average.
Thus the the result is significant
